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

ID: 2051964 • Letter: I

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 3.50 m/s^2 upward. At 25.0 s after launch, the second stage fires for 10.0 s, which boosts the rocket's velocity to 132.5 m/s upward at 35.0 s after launch. This firing uses up all the fuel, however, so after the second stage has finished firing, the only force acting on the rocket is gravity. Air resistance can be neglected.
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 c) for this part you can just use all your already found values for the stage 2 rocket and solve for Vfinal using Vfinal^2 = Vstart^2 + 2 (a)(d) Vfinal^2 = Vstart^2 + 2 (a)(d) Vfianl^2 = 0 + 2(9.8)(1988.125) Vfinal^2 = 38967.25 Vfinal = (square root) 38967.25 Vfinal = 197.4 m/s There you go :) Enjoy! Source(s): Grade 12 Physics ;)

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