Projection problem?

Question

A projection is launched from the ground with speed of 20m/s at the angle 53 degrees above the horizontal towards a high vertical wall 26m away. How high in meters above the ground does the projection strike the wall?

Is it 6.7 meters?

How long was it in the air?

Is it 1.6 seconds?

Answer 1

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Answer 2

First, calculate the initial vertical and horizontal components of velocity. If the magnitude of velocity is v and the angle above the horizontal is θ, the initial vertical velocity v_y and horizontal velocity v_x are found as follows.

v_y0 = v*sin(θ)
v_x0 = v*cos(θ)

From the givens, v = 20 m/s and θ = 53 degrees.

The amount of time the projectile is in the air would be equal to (26 m) / v_x0, because that is the time it would take the projectile to travel the 26 m horizontal distance. I get 1.6 s, the same answer as you.

The height of the projectile at time t is y = v_y0*t - 0.5gt^2, where g = 9.8 m/s^2. To find the height at which the projectile hits the wall, use t = 1.6 s, the value we just calculated for the time when the projectile hits the wall. I get y = 13.01 m, which was not your result.

Answer 3

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