A rocket is fired at a speed of 70.0 m/s from ground level, at an angle of 40.0 ° above the horizontal. The rocket is fired toward an 14.0-m high wall, which is located 21.0 m away. The rocket attains its launch speed in a negligibly short period of time, after which its engines shut down and the rocket coasts. By how much does the rocket clear the top of the wall? I keep getting 1.64 m but its not the correct answer

Question

A rocket is fired at a speed of 70.0 m/s from ground level, at an angle of 40.0 ° above the horizontal. The rocket is fired toward an 14.0-m high wall, which is located 21.0 m away. The rocket attains its launch speed in a negligibly short period of time, after which its engines shut down and the rocket coasts. By how much does the rocket clear the top of the wall? I keep getting 1.64 m but its not the correct answer

Answer 1

t=x/v(0x) = x/v(0) •cosα=

=21/70•cos40°=0.39 s.
h=v(oy) •t - gt²/2 =
=v(o) •sinα•t - gt²/2=
=70•sin40°•0.39 – 9.8•0.39²/2 =16.8 m
Δh=16.8-14 = 2.8 m.

Answer 2

apparently that is incorrect as well according to the assignment program:/

Answer 3

To determine how much the rocket clears the top of the wall, we can break down the motion of the rocket into vertical and horizontal components.

First, let's find the time it takes for the rocket to reach its maximum height. We'll assume the acceleration due to gravity is -9.8 m/s².

Using the vertical motion equation:
vf = vi + at, where vf = 0 m/s (at the highest point), vi = 70.0 m/s (initial vertical velocity), a = -9.8 m/s² (acceleration due to gravity), and t is the time.

0 = 70.0 - 9.8t
9.8t = 70.0
t = 70.0 / 9.8
t ≈ 7.14 s

Now, let's find the height reached by the rocket.

Using the vertical motion equation:
Δy = vit + (1/2)at², where Δy is the change in height, vi = 70.0 m/s, t = 7.14 s, and a = -9.8 m/s².

Δy = 70.0(7.14) + (1/2)(-9.8)(7.14)²
Δy ≈ 338.06 m

So, the rocket reaches a height of approximately 338.06 m.

Next, we'll calculate the horizontal distance the rocket travels.

Using the horizontal motion equation:
Δx = vixt, where Δx is the horizontal distance traveled, vix = 70.0 m/s (initial horizontal velocity), and t = 7.14 s.

Δx = 70.0 * 7.14
Δx ≈ 499.8 m

Now, we can determine the vertical distance from the top of the wall by subtracting the height of the wall (14.0 m) from the maximum height reached by the rocket (338.06 m) and add the height of the wall.

Vertical distance = (338.06 - 14.0) + 14.0
Vertical distance ≈ 338.06 m

Therefore, the rocket clears the top of the wall by approximately 338.06 meters.