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.
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?
3 answers
23m is wrong
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.
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.
0 answers