For simplicity, assume g = constant over the trajectory. This assumes the height change is a small fraction of the earth's radius, 6000 km.
(a) First get the height (H' = 525 m) and time T at engine cutoff.
a*T^2/2 = 525
T = 21.6 s
Velocity at cutoff = Vm = a T = 48.6 m/s
Max height Hm = H' + additional height gained while "coasting" upwards:
Hm = 525 m + Vm^2/(2g)= 645.5 m
(b) Let additional time T'to "crash".
Solve g*T' = Vm, for T'.
To compute speed V" at impact:
V"^2/2 = Vm^2/2 + gH'
a 7500 kilogram rocket blasts off vertically upward from the launch pad with a constant acceleration of 2.25 m/s squared and feels no appreciable air resistance, when it reaches the height of 525 meter its engine suddenly fail so that the only force acting on it is now gravity. (a.) what is the maximum height the rocket will reach above the launch pad? (b.) how much time after engine failure will elapse before the rocket comes crashing down and how fast will it be moving just before it crashes?
2 answers
536 miles