A projectile is launched horizontally at a speed of 40 meters per second from a platform located a vertical distance h above the ground. The projectile strikes the ground after time at horizontal distance from the base of the platform. [Neglect friction.]

Calculate the vertical distance, h, if the projectile's total time of flight is 4.5 seconds. You must show work to earn full credit with the minimum 3 steps:
1. Equation used
2. Substitution with correct units
3. Final answer with correct units

A Projectile Is Launched Horizontally At A Speed Of 40 Meters Per Second From A Platform Located A Vertical

Answers

Answer 1

Answer:1

Explanation:t=rad2h/g

Answer 2

The launch of projectiles allows to find the height of the body launched horizontally is:

The height is 99.2 m

Projectile launching is an application of kinematics where there is no acceleration on the x-axis and the y-axis is the gravity acceleration.

In the attachment we see a diagram of the movement, as the body is thrown horizontally, its initial vertical speed is zero.

           y = y₀ + [tex]v_o_y[/tex] t - ½ g t²

Where y, y₀i are the current and initial heights, respectively, [tex]v_{oy}[/tex] the vertical initial velocity, g the acceleration of gravity and t the time.

           0 =y₀ + 0 - ½ g t²

           y₀ = ½ g t²

let's calculate

           y₀ = ½ 9.8 4.5²

           y₀ = 99.2 m

          y₀ = h = 99.2 m

This is the initial height of the object when it is thrown.

In conclusion using the projectile launch we can find the height of the horizontally launched body is:

The height is 99.2 m

Learn more here:  brainly.com/question/10903823

A Projectile Is Launched Horizontally At A Speed Of 40 Meters Per Second From A Platform Located A Vertical

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1. A plane flying horizontally at an altitude of 490 meters and having a velocity of 250 meters per second east, drops a supply packet to a work crew on the ground. It falls freely without a parachute. [Assume no wind and negligible air resistance.] A. Calculate the time required for the packet to hit the ground. (25 pts)

Answers

Answer:

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Explanation:

Given :

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Answers

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Answers

Answer:

Explanation:

From the given information:

The car's initial velocity = 4 m/s in the direction of east 45° due north

We can therefore express the vector of this component form as:

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The magnitude of the avg. acceleration is:

[tex]|| a||= \sqrt{(0.370 m/s^2)^2 + (-0.711 m/s^2)^2)[/tex]

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i.e.

[tex]tan(\theta) = \dfrac{-0.711 m/s^2}{ 0.370 m/s^2}[/tex]

[tex]tan(\theta) = -1.9216[/tex]

[tex]\theta = tan^{-1} ( -1.9216 )[/tex]

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Answers

Answer:

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Answers

Answer:

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Answers

Answer:

height = 45 m

Explanation:

Initial velocity = 0 m/s

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[tex] {v}^{2} - {u}^{2} = 2gh[/tex]

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Answers

Answer:

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Answers

Answer:

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Explanation:

Given parameters:

Number of turns in primary coil  = 1000 turns

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Unknown:

Number of turns in secondary coil = ?

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To solve this problem, we have to use the expression for solving problems related with transformers;

   

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          46 x 1000 = 230Ns

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