Given :
A kayak on a lake is moving at a constant speed of 4.1 m/s in a direction 41.6° north of west.
To Find :
After 10 min :
(a) how far west the kayak has traveled km.
(b) how far north the kayak has traveled km.
Solution :
a) Time taken in seconds :
[tex]t=10\times 60 \ s = 600\ s[/tex] .
Distance travelled in 600 s is :
[tex]D=s\times t\\\\D=4.1\times 600\ m\\\\D=2460\ m[/tex]
Distance in km is :
[tex]D=\dfrac{2460}{1000}=2.46\ km[/tex]
b) Now , distance from the north is given by :
[tex]d=D\times sin\ 41.6^o\\\\d=2.46 \times sin\ 41.6^o\\\\d=1.63\ km[/tex]
Hence , this is the required solution .
A car stops in 120 m. If it has an acceleration of –5m/s 2 , how long did it take to stop
Answer:
t=240s
Explanation:
Distance=120m
Acceleration=-5m/s^2
v=0
Let u=x m/s
Using equation v^2-u^2=2as:-
0-x=2(-5)(120)
-x=-1200
x=1200m/s
Using now equation v=u+at:-
0=1200+-5t
5t=1200
t=240s
If a car stops at 120 meters. If it has an acceleration of –5 meters/second², then it would take 6.928 seconds to stop.
What is acceleration?The rate of change of the velocity with respect to time is known as the acceleration of the object.
As given in the problem a car stops at 120 meters. If it has an acceleration of –5 meters/second², then we have to find out how long it would take seconds to stop.
By using the second equation of motion,
s = ut + 1/2at²
The distance traveled by car before stopping = 120 meters
acceleration = –5 meters/second²
-120 = 0 + 0.5×( –5)t²
t² = 120/2.5
t² =48
t = 6.928 seconds
Thus, the time taken by the car before stopping would be 6.928 seconds.
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When charges qa, qb, and qc are placed respectively at the corners a, b, and c of a right triangle, the potential at the midpoint of the hypotenuse is 20 V. When the charge qa is removed, the potential at the midpoint becomes 15 V. When, instead, the charge qb is removed (qa and qc both in place), the potential at the midpoint becomes 12 V. What is the potential at the midpoint if only the charge qc is removed from the array of charges?
Answer:
8v
Explanation:
First we apply super position principle
Vt= v1 + v2+ v3
Remove qa
But vt= 20v
So V = v2+v3
V1= 20-15
= 5v
Remove qb
V= v1+v3
V=8v
So the potential when qa and qc are remove is the potential due to qb
Which is 8v
Assuming it is a van der Waals gas, calculate the critical temperature, pressure and volume for CO2. (a = 3.610 atm L2 mol-2, b = 0.0429 L mol-1)
pc = ___ atm
Tc = ___ K
Vc = ___ L/mol
Answer
To get critical pressure
We use
Pc = a/(27b²)
So
= 3.610/(27 X 0.0429²)
We have
= 72.7 atm
Critical temperaturewe
We use
Tc = 8a/27Rb
= 8 x 3.610/(27 x 0.0812 x 0.0429)
= 307 K
Critical volume
We use
Vc =3b =
3 x 0.0429
= 0.129L/mol
4. Lead has a density of 11.5g/cmº. A rectangular block of lead measures 7cm x5cmx2cm.
a) Find the volume of the block of lead.
b) Find the mass of the block of lead
Answer:
(a) 70cm³
(b) 805 grams
Explanation:
(a) V = L×B×H
= 7cm×5cm×2cm
= 35cm×2cm
= 70cm³
(b) Mass = Volume × Density
= 70cm³ × 11.5g/cm³
= 805 grams
A -5.40nC point charge is on the x axis at x = 1.25m . A second point charge Q is on the x axis at -0.625m.
A) What must be the charge Q for the resultant electric field at the origin to be 50.0N/C in the +x direction?
B) What must be the charge Q for the resultant electric field at the origin to be 50.0N/C in the -x direction?
Answe
a) Q = 0.820 10⁻⁹ C , b) Q = -3.52 10⁻⁹ C
Explanation:
The electric field is given by the formula
E = k q / r²
where E is a vector quantity, so it must be added as a vector
E_total = E₁ + E₂
let's look for the two electric fields
E₁ = k q₁ / r₁²
E₁ = 9 10⁹ 5.4 10⁻⁹ / 1.25²
E₁ = 31.10 N / C
E2 = k Q / r₂²
E2 = 9 10⁹ Q / 0.625²
E2 = 23.04 10⁹ Q N / C (1)
now we can solve the two cases presented
a) The total field is
E_total = 50.0 N / C towards + x
since the test charge is positive the electric field E1 points to the right in the direction of the + x axis, so the equation is
E_total = E1 + E₂
E₂ = E_toal - E₁
E₂ = 50.0 -31.10
E2 = 18.9 N /C
With the value of the electric field we can calculate the charge (Q) using equation 1
E₂ = 23.04 10⁹ Q
Q = E₂ / 23.04 10⁹
Q = 18.9 / 23.04 10⁹
Q = 0.820 10⁻⁹ C
the charge on Q is positive
b) E_total = -50.0 N / C
E_total = E₁ + E₂
E₂ = E_total - E₁
E2 = -50.0 - 31.10
E2 = -81.10 N /C
we calculate the charge
Q = E2 / 23.04 10⁹
Q = -81.1 / 23.04 10⁹
Q = -3.52 10⁻⁹ C
for this case the charge is negative
Which scientist determined that electrons had predicted zones orbiting the nucleus?
Answer:
Rutherford
Explanation:
Because
Schrödinger I just took the unit review
Two 100kg bumper cars are moving towards eachother in oppisite directions. Car A is moving at 8 m/s and Car B at -10 m/s when they collide head on. If the resulting velocity of Car B after the collision is 8 m/s, what is the velocity of Car A after the collision
Answer:
[tex]-10 m/s[/tex]
Explanation:
When two cars collide then the momentum of two cars will remains conserved
Mass of two cars = 100 kg Speed of car A = 8 m/s Speed of car B = - 10 m/s After collision the speed of car B = +8 m/sBy momentum conservation equation
[tex]m1v1i+m2v2i=m1v1f + m2v2f[/tex]
[tex](100)(8)+(100)(-10)=(100v)+(100)(8)\\ v=-10 m/s[/tex]
A bungee jumper with mass 65.0 kg jumps from a high bridge. After reaching his lowest point, he oscillates up and down, hitting a low point eight more times in 38.0 s. After many oscillations, he finally comes to rest 25.0 m below the level of the bridge. Calculate the spring stiffness constant and the unstretched length of the bungee cord.
Explanation:
It is given that,
Mass of a bungee jumper is 65 kg
The time period of the oscillation is 38 s, hitting a low point eight more times.It means its time period is
[tex]T=\dfrac{38}{8}\\\\T=4.75\ s[/tex]
After many oscillations, he finally comes to rest 25.0 m below the level of the bridge.
For an oscillating object, the time period is given by :
[tex]T=2\pi \sqrt{\dfrac{m}{k}}[/tex]
k = spring stiffness constant
So,
[tex]k=\dfrac{4\pi ^2m}{T^2}\\\\k=\dfrac{4\pi ^2\times 65}{(4.75)^2}\\\\k=113.43\ N/m[/tex]
When the cord is in air,
mg=kx
x = the extension in the cord
[tex]x=\dfrac{mg}{k}\\\\x=\dfrac{65\times 9.8}{113.6}\\\\x=5.6\ m[/tex]
So, the unstretched length of the bungee cord is equal to 25 m - 5.6 m = 19.4 m
The spring stiffness constant is 116.7 N/m and the the unstretched length of the bungee cord is 19.54 m.
The given parameters;
mass of the bungee jumper, m = 65 kgtime of motion, t = 38 sdistance to come to rest, d = 25 mThe period of oscillation of the bungee jumper is calculated as follows;
[tex]T = \frac{t}{n} \\\\T = \frac{38}{8} \\\\T = 4.75 \ s[/tex]
The spring stiffness constant is calculated as follows;
[tex]T = 2\pi \sqrt{\frac{m}{k} } \\\\\sqrt{\frac{m}{k} } = \frac{T}{2\pi} \\\\k = m \times \frac{T^2}{4\pi^2} \\\\k = 65 \times \frac{(4.75)^2}{4\pi ^2} \\\\k = 116.7 \ N/m[/tex]
The extension of the cord is calculated as follows;
[tex]F = kx\\\\mg = kx\\\\x = \frac{mg}{k} \\\\x = \frac{65 \times 9.8}{116.7} \\\\x = 5.46 \ m[/tex]
The unstretched length of the bungee cord is calculated as;
[tex]\Delta x = l_2-l_1\\\\l_1 = l_2 - \Delta x\\\\l_1 = 25 - 5.46\\\\l_1 = 19.54 \ m[/tex]
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2. If a cyclist in the Tour de France traveled southwest a distance of 12,250 meters in one hour, what would the velocity of the cyclist be?
12,25 km/h
≈ 3,4 m/s
Explanation:v = d/t
= 12250m/h
= 12,25km/h
or
v = d/t
= 12250m/h
1h = 60m×60s = 3600s
= 12250m/3600s
≈ 3,4 m/s
A Lotus will travel 275 meters in 4.71 seconds. What is this car's average speed?
A stone is thrown vertically upward with a speed of 28.0 m/s how much time is required to reach this height
How does sleep affect your ability to handle stress?
Answer: Stress can adversely affect sleep quality and duration, while insufficient sleep can increase stress levels. Both stress and a lack of sleep can lead to lasting physical and mental health problems.
Explanation:
Many report that there stress increases when the length and quality of their sleep decreases. When you do not get enough sleep, 21 percent of adults report feeling more stressed.
The speed of a bus increases uniformly from
15 ms- to 60 ms in 20 seconds. Calculate
a. the average speed,
b. the acceleration,
C. the distance travelled during the entire
period The speed of a bus increases uniformly from
15 ms- to 60 ms in 20 seconds. Calculate
a. the average speed,
b. the acceleration,
C. the distance travelled during the entire
period
Explanation:
a. For constant acceleration:
v_avg = ½ (v + v₀)
v_avg = ½ (60 m/s + 15 m/s)
v_avg = 37.5 m/s
b. a = (v − v₀) / t
a = (60 m/s − 15 m/s) / 20 s
a = 2.25 m/s²
c. x = v_avg t
x = (37.5 m/s) (20 s)
x = 750 m
which water molecules have the greatest kinetic energy
A ball with a mass of 3.7 kg is thrown downward with an initial velocity of 8 m/s from a high building. How fast will it be moving after 3 seconds?
Answer:
v=37.4 m/s
Explanation:
It is given that,
Mass of a ball, m = 3.7 kg
Initial velocity of the ball is u = 8 m/s
We need to find its velocity after 3 seconds. It is moving downwards. The equation of motion is this case is
v=u+gt
[tex]v=8+9.8\times 3\\\\v=37.4\ m/s[/tex]
So, the velocity of the ball after 3 seconds is 37.4 m/s.
Two pounds of water vapor at 30 psia fill the 4-ft3 left chamber of a partitioned system. The right chamber has twice the volume of the left and is initially evacuated. Determine the pressure of the water after the partition has been removed and enough heat has been transferred so that the temperature of the water is 40oF.
Answer:
3.38atm
Explanation:
Using data from the steam table we have that
Moles of water vapour = 907.19 / 18
= 50.4 moles
So
p1 = 30 psi = 30 x 0.68 = 2.04 atm
v1 = 4ft³= 113.2 L
Then from
PV= nRT
Then to find T we use
T1 = p1 V1 / n R
= 2.04 x 113.2 / 50.4 x 0.0821
= 55.8 K
Then to find volume two
v2 = 2v1 + v1
So
3 v1 = 339.6 K
The pressure two we use
P2 = n R T2 / V2
= 50.4 x 0.0821 x 277.6 / 339.6
So we have
= 3.38 atm =
A ball of mass m moving with speed V collides with another ball of mass 2m (e= 1/2) in a horizontal smooth fixed circular tube of radius R (R is sufficiently large R>>>d). The time after which next collision will take place is:________
Answer:
[tex]$ \frac{4\pi R}{V}$[/tex]
Explanation:
Given :
Mass of ball 1 = m
Mass of ball 2 = 2m
Since, R>>>d, the collision is head on.
Therefore, we get
[tex]$ \frac{v_1 -v_2}{V}=\frac{1}{2}$[/tex]
[tex]$ \therefore \frac{\text{velocity of seperation}}{\text{velocity of approach}}= v_1-v_2 = \frac{V}{2}$[/tex]
Relative velocity is given by V/2. So, we get the time when the masses will again collide as
[tex]$ t = \frac{2\pi R}{\frac{V}{2}}=\frac{4\pi R}{V} $[/tex]
Finally, consider the expression (6.67 x 10^-11)(5.97 x 10^24)/(6.38 x 10^6)^2 Determine the values of a and k when the value of this expression is written in scientific rotation. Enter a and k, separated by commas.
Explanation:
We need to find the value of following expression :
[tex]\dfrac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{(6.38 \times 10^6)^2}[/tex]
Firstly, solving the numerator of the above expression :
[tex]=\dfrac{39.8199\times 10^{-11+24}}{40.7044\times 10^{12}}\\\\=\dfrac{39.8199\times 10^{13}}{40.7044\times 10^{12}}\\\\=9.7827[/tex]
Rounding off the result = 9.78
In scientific notation : [tex]9.78\times 10^0[/tex]
The value of a = 9.78 and k = 0.
A satellite dish has the shape of a parabola when viewed from the side. The dish is inches wide and inches deep. How far is the receiver from the bottom of the dish if the receiver is placed at the focus
Complete question is;
A satellite dish has the shape of a parabola when viewed from the side. The dish is 60 inches wide and 45 inches deep. How far is the receiver from the bottom of the dish if the receiver is placed at the focus?
Answer:
the receiver should be put 40 inches from the bottom of the dish on the concave side of the dish
Explanation:
The base of the dish would simply be the vertex of parabola.
Since we want to find how far the receiver is from the bottom, the place where we'll place the receiver is simply the focus of the parabola.
Now, for example, if this is a parabola that opens upward and has it's vertex at the origin, then half of the diameter at a height of 45 inches gives the two points (60, 22.5) and (-60, 22.5)
Standard form equation of parabola with vertex at origin and pointing upwards is given by;
x² = 4ay
Plugging in the values of x and y gives;
60² = 4a(22.5)
3600/90 = a
a = 40 inches
Thus, the receiver should be put 40 inches from the vertex on the concave side of the dish
A commuter backs her car out of her garage with an acceleration of 1.40 m/s2.A) How long does it take her to reach a speed of 2.00 m/s?B) If she then breaks to a stop in 0.800 s, what is her deceleration?
Answer:
(A) 1.43secs
(B) -2.50m/s^2
Explanation:
A commuter backs her car out of her garage with an acceleration of 1.40m/s^2
(A) When the speed is 2.00m/s then, the time can be calculated as follows
t= Vf-Vo/a
The values given are a= 1.40m/s^2 , Vf= 2.00m/s, Vo= 0
= 2.00-0/1.40
= 2.00/1.40
= 1.43secs
(B) The deceleration when the time is 0.800secs can be calculated as follows
a= Vf-Vo/t
= 0-2.00/0.800
= -2.00/0.800
= -2.50m/s^2
(7) A 2500 lbm car moving at 25 mi/hr is accelerated at a constant rate of 15 ft/s2 up to a speed of 50 mi/hr. What is the force and total time required?
Answer:
The force is [tex]F = 1164.6\ lbf[/tex]
The time is [tex]\Delta t = 2.44 \ s[/tex]
Explanation:
From the question we are told that
The mass of the car is [tex]m = 2500 \ lbm[/tex]
The initial velocity of the car is [tex]u = 25 \ mi/hr[/tex]
The final velocity of the car is [tex]v = 50 \ mi/hr[/tex]
The acceleration is [tex]a = 15 ft/s^2 = \frac{15 * 3600^2}{ 5280} = 36818.2 \ mi/h^2[/tex]
Generally the acceleration is mathematically represented as
[tex]a = \frac{v-u}{\Delta t}[/tex]
=> [tex]36818.2 = \frac{50 - 25 }{ \Delta t}[/tex]
=> [tex]t = 0.000679 \ hr[/tex]
converting to seconds
[tex]\Delta t = 0.0000679 * 3600[/tex]
=> [tex]\Delta t = 2.44 \ s[/tex]
Generally the force is mathematically represented as
[tex]F = m * a[/tex]
=> [tex]F = 2500 * 15[/tex]
=> [tex]F = 37500 \ \frac{lbm * ft}{s^2}[/tex]
Now converting to foot-pound-second we have
[tex]F = \frac{37500}{32.2}[/tex]
=> [tex]F = 1164.6\ lbf[/tex]
Explain why a Merry-Go-Round and a Ferris Wheel have a constant acceleration when they are moving?
Answer:
merry go round and Ferris wheel have a constant acceleration due to the change in direction at every point.
Answer:
A merry-go-round is accelerating. Acceleration is a change in speed, direction, or both. Even though the speed of the merry-go-round does not change, its direction constantly changes as it spins.
Explanation:
An 85 kg skydiver is falling through the air at a constant speed of 195 km h-1. At what rate does air resistance remove energy from the skydiver?
Answer:
46041J
Explanation:
Using Energy lost= mgh
Changing to standard its we have
= 195*1000/3600=54.2m/s
So = 85*54.2*10= 46041J
Answer:
45167.15 J/s
Explanation:
mass of the man = 85 kg
The man's speed = 195 km/h = 195 x 1000/3600 = 54.167 m/s
The man's weight = mg
where
m is the mass
g is acceleration due to gravity = 9.81 m/s^2
weight = 85 x 9.81 = 833.85 N
The rate at which energy is removed from the man = speed x weight
==> 54.167 x 833.85 = 45167.15 J/s
What happens to the temperature as altitude increases in the exosphere? Does it increase or decrease the higher it goes?
Answer:
it gets colder the higher you go
A paper airplane is thrown horizontally with a velocity of 20 mph. The plane is in the air for 7.63 s before coming to a standstill on the ground. What is the acceleration of the plane?
Answer:
-1.17 m/s²
Explanation:
Given:
v₀ = 20 mph = 8.94 m/s
v = 0 m/s
t = 7.63 s
Find: a
v = at + v₀
0 m/s = a (7.63 s) + 8.94 m/s
a = -1.17 m/s²
The acceleration of the plane will be:
"-1.17 m/s²".
Acceleration and VelocityAccording to the question,
Velocity, v₀ = 20 mph or,
= 8.94 m/s
and,
v = 0 m/s
Time, t = 7.63 s
We know the relation,
→ v = at + v₀
By substituting the values,
0 = a × 7.63 + 8.94
7.63a = - 8.94
a = -[tex]\frac{8.94}{7.63}[/tex]
= - 1.17 m/s²
Thus the response above is correct.
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Public television station KQED in San Francisco broadcasts a sinusoidal radio signal at a power of 777 kW. Assume that the wave spreads out uniformly into a hemisphere above the ground. At a home 5.00 km away from the antenna,
(a) what average pressure does this wave exert on a totally reflecting surface,
(b) what are the amplitudes of the electric and magnetic fields of the wave, and
(c) what is the average density of the energy this wave carries?
(d) For the energy density in part (c), what percentage is due to the electric field and what percentage is due to the magnetic field?
Answer:
A) P = 3.3 × 10^(-11) Pa
B) Amplitude of electric field = 1.931 N/C
Amplitude of magnetic field = 6.44 × 10^(-9) T
C) μ_av = 1.65 × 10^(-11) J/m³
D) 50% each for the electric and magnetic field
Explanation:
A) First of all let's calculate intensity.
I = P_av/A
We are given;
P_av = 777 KW = 777,000 W
Distance = 5 km = 5000 m
Thus;
I = 777000/(2π × 5000²)
I = 0.00495 W/m²
Now, the average pressure would be given by the formula;
P = 2I/C
Where C is speed of light = 3 × 10^(8) m/s
P = (2 × 0.00495)/(3 × 10^(8))
P = 3.3 × 10^(-11) Pa
B) Formula for the amplitude of the electric field is gotten from;
E_max = √[2I/(εo•c)].
Where εo is the Permittivity of free space with a constant value of 8.85 × 10^(−12) c²/N.mm²
I and c remain as before.
Thus;
E_max = √[(2 × 0.00495)/(8.85 × 10^(−12) × 3 × 10^(8))]
E_max = √3.72881355932
E_max = 1.931 N/C
Formula for amplitude of magnetic field is gotten from;
B_max = E_max/c
B_max = 1.931/(3 × 10^(8))
B_max = 6.44 × 10^(-9) T
C) Formula for average density is;
μ_av = εo(E_rms)²
Now, E_rms = E_max/√2
Thus;
E_rms = 1.931/√2
μ_av = 8.85 × 10^(−12) × (1.931/√2)²
μ_av = 1.65 × 10^(-11) J/m³
D) The energy density for the electric and magnetic field is the same. So both of them will have 50% of the energy density.
if a cart goes around a turn at 20 km/h ,what remains constant
1.position
2.velocity
3.direction
4.speed
Answer: 4.speed
Explanation:
In this case, we know that the cart remains at a constant 20km/h.
Now, one could say that "the velocity remains constant, because it always is 20km/h"
But remember that velocity is a vector, so this has a direction, and if the cart is going around a turn, then the direction of motion is changing, which tell us that there is acceleration.
But the module of the velocity, the speed, remains constant at 20km/h.
Then the correct option is 4, speed.
A ski lift has a one-way length of 1 km and a vertical rise of 200 m. The chairs are spaced 20 m apart, and each chair can seat three people. The lift is operating at a steady speed of 10 km/h. Neglecting friction and air drag, and assuming that the average mass of each loaded chair is 250 kg, determine the power required to operate this ski lift. Also, estimate the power required to accelerate this ski lift in 17 s to its operating speed when it is first turned on.
Answer:
Explanation:
The question states that the chairs are spaced 20 m apart through a length of 1 km, or say, 1000 m.
It also does say that each chair weighs 250 kg, and as such the load is
M = 50 * 250
M = 12500.
Taking into consideration, the initial and final heights, we have
h1 = 0, h2 = 200 m
The work needed to raise the chairs,
W = mgh, where h = h2 - h1
W = 12500 * 9.81 * (200 - 0)
W = 2.54*10^7 J
The work is done at a rate of 10 km/h, and at a distance of 1 km, time taken would be
t = 1/10 = 0.1 h or say, 360 s
The power needed thus, is
P = W/t
P = 2.54*10^7 / 360
P = 68125 W, or 68 kW
Initial velocity, u = 0 m/s
Final velocity, v = 10 km/h = 2.78 m/s
Startup time, t is 17 s
Acceleration during the startup then is
a = (v - u)/t
a = 2.78/17
a = 0.163 m/s²
The power needed for the acceleration is
P = ½m [(v² - u²)/t]
P = ½ * 12500 * [2.78²/17]
P = 6250 * 0.455
P = 2844 W
If one object (a) is moving at 60m/s^2, and the other object (b) is moving at 65m/s^2, at what time will the faster moving object be 10m ahead of the other object?
Answer:
a is moving at 60m and the other object
Gwen releases a rock at rest from the top of a 40-m tower. If g = 9.8 m/s2 and air resistance is negligible, what is the speed of the rock as it hits the ground?
Answer:
[tex]28\; \rm m \cdot s^{-1}[/tex].
Explanation:
Short ExplanationApply the SUVAT equation [tex]\left(v^2 - u^2) = 2\, a \, x[/tex], where:
[tex]v[/tex] is the final velocity of the object,[tex]u[/tex] is the initial velocity of the object, [tex]a[/tex] is the acceleration (should be constant,) and[tex]x[/tex] is the displacement of the object while its velocity changed from [tex]v[/tex] to [tex]u[/tex].Assume that going downwards corresponds to a positive displacement. For this question:
[tex]v[/tex] needs to be found.[tex]u = 0[/tex] because the rock is released from rest.[tex]a = g = 9.8 \; \rm m\cdot s^{-2}[/tex].[tex]x = 40\; \rm m[/tex].Solve this equation for [tex]v[/tex]:
[tex]\displaystyle v = \sqrt{2\, a\, x + u^2} = \sqrt{2\times 9.8 \times 40} = 28\; \rm m \cdot s^{-1}[/tex].
In other words, the rock reached a velocity of [tex]28\; \rm m\cdot s^{-1}[/tex] (downwards) right before it hits the ground.
ExplanationLet [tex]v[/tex] be the velocity (in [tex]\rm m \cdot s^{-1}[/tex]) of this rock right before it hits the ground. Under the assumptions of this question, it would take a time of [tex]t = (v / 9.8)[/tex] seconds for this rock to reach that velocity if it started from rest and accelerated at [tex]9.8\; \rm m \cdot s^{-2}[/tex].
Note that under these assumptions, the acceleration of this rock is constant. Therefore, the average velocity of this rock would be exactly one-half the sum of the initial and final velocity. In other words, if [tex]u[/tex] denotes the initial velocity of this rock, the average velocity of this rock during the fall would be:
[tex]\displaystyle \frac{u + v}{2}[/tex].
On the other hand, [tex]u = 0[/tex] because this stone is released from rest. Therefore, the average velocity of this rock during the fall would be exactly [tex](v / 2)[/tex].
The displacement of an object over a period of time is equal to the length of that period times the average velocity over that period. For this rock, the length of this fall would be [tex]t = (v / 9.8)[/tex], while the average velocity over that period would be [tex](v / 2)[/tex]. Therefore, the displacement (in meters) of the rock during the entire fall would be:
[tex]\displaystyle \left(\frac{v}{2}\right) \cdot \left(\frac{v}{9.8}\right) = \frac{v^2}{19.6}[/tex].
That displacement should be equal to the change in the height of the rock, [tex]40\; \rm m[/tex]:
[tex]\displaystyle \frac{v^2}{19.6} = 40[/tex].
Solve for [tex]v[/tex]:
[tex]v = 28\; \rm m \cdot s^{-1}[/tex].
Once again, the speed of the rock would be [tex]28\;\rm m \cdot s^{-1}[/tex] right before it hits the ground.