Answer to Question #77990 in Physics for Janice

Question

A rocket moves straight upward, starting from rest with an acceleration of +26.5 m/s2. It runs out of fuel at the end of 4.92 s and continues to coast upward, reaching a maximum height before falling back to Earth.

Find the rocket's velocity and position at the end of 4.92 s.

vb = _______ m/s
yb = ________ m


Find the maximum height the rocket reaches.

Find the velocity the instant before the rocket crashes on the ground.

Answer 1

The rocket's velocity and position at the end of 4.92 s:
v_b=at=(26.5)(4.92)=130 m/s.
y_b=1/2 at^2=0.5(26.5) (4.92)^2=320 m.
The maximum height the rocket reaches:
H=y_b+(v_b^2)/2g=320+〖130〗^2/2(9.8) =1180 m.
The velocity the instant before the rocket crashes on the ground:
V=√2gH=√(2(9.8)1180)=152 m/s.