A rocket is fired at a speed of 75.0 m/s from ground level, at an angle of 61.2° above the horizontal. The rocket is fired toward an 11.0 m high wall, which is located 20.5 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?

Question

A rocket is fired at a speed of 75.0 m/s from ground level, at an angle of 61.2° above the horizontal. The rocket is fired toward an 11.0 m high wall, which is located 20.5 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?

Answer 1

We will neglect air resistance and propulsion of the rocket, and treat the rocket as a projectile.

v0=initial velocity = 75 m s-1
θ=angle with horizontal

horizontal component of velocity, vh
=v0cos θ
Distance from the wall, D = 20.5 m
Time to reach the wall, t1
= D/vh

Vertical component of initial velocity
= vv
= v sin θ

Height h of rocket after time t1
= vv*t1 -g (t1)²

Clearance from the wall, C
= h - 11m

I get about 23m.

Answer 2

23m is wrong

Answer 3

Thank you Brittany, there was a factor (1/2) missing near the end. Here is the corrected calculation.

Height h of rocket after time t1
= vv*t1 -(1/2)g(t1)²

Clearance from the wall, C
= h - 11m

I get a little less than 25m using g=9.8 m s-2.

Answer 4

To find out by how much the rocket clears the top of the wall, we need to calculate the height the rocket reaches and compare it with the height of the wall.

Let's break down the problem into horizontal and vertical components.

First, let's find the time it takes for the rocket to reach the wall. We can use the horizontal component of the rocket's velocity for this calculation.

The horizontal component of the velocity (Vx) can be calculated using the initial velocity (75.0 m/s) and the angle of projection (61.2°).

Vx = initial velocity * cos(angle)
Vx = 75.0 m/s * cos(61.2°)
Vx = 75.0 m/s * 0.5008 (rounded to 4 decimal places)
Vx = 37.56 m/s

Now, let's calculate the time it takes for the rocket to reach the wall.

Distance = velocity * time

Since the distance covered is given (20.5 m) and the horizontal velocity is known (37.56 m/s), we can rearrange the equation to solve for time:

time = distance / velocity
time = 20.5 m / 37.56 m/s
time = 0.546 seconds (rounded to 3 decimal places)

Now that we have the time it takes for the rocket to reach the wall, let's find the maximum height it reaches.

The vertical component of the velocity (Vy) can be calculated using the initial velocity (75.0 m/s) and the angle of projection (61.2°).

Vy = initial velocity * sin(angle)
Vy = 75.0 m/s * sin(61.2°)
Vy = 75.0 m/s * 0.8609 (rounded to 4 decimal places)
Vy = 64.57 m/s

We can use this vertical velocity and the time calculated earlier to find the maximum height (H) reached by the rocket using the following equation:

H = (Vy * time) - (0.5 * g * time^2)

Where g is the acceleration due to gravity, approximately 9.8 m/s^2.

H = (64.57 m/s * 0.546 s) - (0.5 * 9.8 m/s^2 * (0.546 s)^2)
H = 35.2 m - 1.524 m
H = 33.676 m (rounded to 3 decimal places)

Finally, we need to find out how much the rocket clears the top of the wall. We subtract the height of the wall (11.0 m) from the maximum height attained by the rocket.

Clearance = Maximum height - Height of the wall
Clearance = 33.676 m - 11.0 m
Clearance = 22.676 m (rounded to 3 decimal places)

Therefore, the rocket clears the top of the wall by approximately 22.676 meters.