A rocket blasts off vertically from rest on the launch pad with an upward accele

Question

A rocket blasts off vertically from rest on the launch pad with an upward acceleration of 2.70 { m m/s^2}. At 10.0 { m s} after blastoff, the engines suddenly fail, which means that the force they produce instantly stops.How high above the launch pad will the rocket eventually go? Find the rocket's velocity at its highest point. Find the magnitude of the rocket's acceleration at its highest point. How long after it was launched will the rocket fall back to the launch pad? How fast will it be moving when it does so?

Explanation / Answer

height = ½at² = ½•2.6m/s²•(30s)² = 1170 m

velocity at highest point = 0 (switching from +ve to -ve)
acceleration at all points after engine failure = -9.8 m/s² (-ve meaning downward)

to fall from max height can be found using
s = 1170m = ½at² = ½•9.8m/s²•t²
t = 15.5s
so total flight time = 45.5 s

falling for 15.5s produces a speed of
v = at = 9.8m/s²•15.5s = 151 m/s

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