Question

A rocket is fired at a speed of 75.0 m/s from ground level, at an angle of 55.1° above the horizontal. The rocket is fired toward an 11.0 m high wall, which is located 28.0 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?

Answer 1

PROJECTILE

1)

along horizontal
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initial velocity v0x = v*costheta

acceleration ax = 0

initial position = xo = 0

final position = x = 28

displacement = x - x0

from equation of motion

x - x0 = v0x*T+ 0.5*ax*T^2

x - x0 = v*costheta*T

T = (x - x0)/(v*costheta)......(1)

along vertical
______________

initial velocity v0y = v*sintheta

acceleration ay = -g = -9.8 m/s^2

initial position y0 = 0

final position y = ?

from equation of motion

y-y0 = v0y*T + 0.5*ay*T^2 .........(2)

using 1 in 2

y-y0 = (v*sin(theta)*(x-x0))/(v*cos(theta)) - (0.5*g*(x-x0)^2)/(v^2*(cos(theta))^2)

y - y0 = (x - x0)*tantheta - (0.5*g*(x-x0)^2)/(v^2*(cos(theta))^2)

y - 0 = 28*tan55.1 - (0.5*9.8*28^2/(75^2*(cos55.1)^2))

y = 38.1 m

the rocket clear the top by 38.1 - 11 = 27.1 m