a model rocket is launched straight upward with an initial speed of 50m/s. iit a

Question

a model rocket is launched straight upward with an initial speed of 50m/s. iit accelerates with a constant upward acceleration of 2.0 m/s^2 until its engine stops at an altitude of 150m. calculate the maximum heigght reached by the rocket and how long the rocket is in the air.

Explanation / Answer

The velocity of the rocket at height 150 m is v^2 = u^2 - 2 a (y - yo) = 50^2 - 2 * 2 * (150-0) = 2500 - 600 = 1900 ==> v = 43.58 m/s The maximum height reached by the rocket y = 150 + v^2/2g = 150 + 96.89 = 246.89 m The time taken to reach height 150 m is (y-yo) = vavg * t 150 = (50 + 43.58)/2 * t ==> t = 3.205 s The time taken to reach 246.89 m is y = yo + vavg * t 246.89 = 150 + (0 + 43.58)/2 * t' ==> t' = 4.446 s total time T = 3.205 + 4.446 = 7.651 s The time taken to reach the groung from height 246.89 m is t'' = [2 y / g] ^1/2 = [2 * 246.89 / 9.8]^1/2 = 7.098 s Therefore total time in air = T + t'' = 7.651 + 7.089 = 14.74s

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