In the first stage of a two-stage rocket, the rocket is fired from the launch pa

Question

In the first stage of a two-stage rocket, the rocket is fired from the launch pad starting from rest but with a constant acceleration of 3.50m/s^2 upward. At 25.0rm s after launch, the rocket fires the second stage, which suddenly boosts its speed to 132.5m/s upward. This firing uses up all the fuel, however, so then the only force acting on the rocket is gravity. Air resistance is negligible.
a)How much time after the stage-two firing will it take for the rocket to fall back to the launch pad?
b)How fast will the stage-two rocket be moving just as it reaches the launch pad?

Explanation / Answer

a) First you use the formula d = Vstart(t) + 1/2(a)(t)^2 to find the distance the 1st rocket brings us. d = Vstart(t) + 1/2(a)(t)^2 d = 0(25) + 1/2 (3.5)(25)^2 d = 1093.75 m Now we know that it will slowly lose speed at a rate of 9.8 m/s due to Gravity. Therefore: t = 132.5 / 9.8 t = 13.5 s At that point the rocket has come to rest. we can now use the equation d = 1/2 (Vstart + Vfinal)t to determine the distance traveled. d = 1/2 (Vstart + Vfinal)t d = 1/2 (132.5 + 0)13.5 d = 894.375 D1 + D2 = 1093.75 + 894.375 = 1988.125 b) now we can simply use the equation d = Vstart (t) + 1/2 (a)(t)^2 to find the time from the highest point to landing. d = Vstart (t) + 1/2 (a)(t)^2 1988.125 = 0(t) + 1/2 (9.8)(t)^2 2(1988.125) / 9.8 = t^2 (square root) 405.739 = t t = 20.1s T1 + T2 = 13.5 +20.1 = 33.6s

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