Thursday
March 23, 2017

Post a New Question

Posted by on Monday, July 19, 2010 at 4:53am.

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?

  • college physics - , Monday, July 19, 2010 at 6:18am

    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'

  • college physics - , Wednesday, July 21, 2010 at 6:27pm

    536 miles

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question