Question #77990

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.

Expert's answer

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.
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