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?
Answers
MathMate
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.
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.
brittany
23m is wrong
MathMate
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 -<b>(1/2)</b>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 -<b>(1/2)</b>g(t1)²
Clearance from the wall, C
= h - 11m
I get a little less than 25m using g=9.8 m s-2.