Question

Carlon launched a toy rocket vertically from ground level (y=0m, time=0s). The rocket's engine provides constant acceleration during the burn phase. At the instant of engine burnout the rocket has risen 89m. The rocket continues to rise in unpowered flight, reaches mex height of 232m, and then falls back to the ground. What is the rocket engine upward acceleration closest too? A. 12.1m/s B. 22.3 m/s C 15.8 m/s

Answer 1

(c)15.8 m/s^2

explanation:

From eqn of motion

vf^2 = vi^2 + 2 a S

At the max height, vf = 0 ; S = 232 - 89 = 143 m ; a = g = -9.8

vi^2 - 2 g S = 0

vi = sqrt ( 2 g S) = sqrt ( 2 x 9.81 x 143) = 53 m/s

The work done by the upward thrust should be equal to the KE gain of the rokcet

W(thrust) = KE(rocket)

(ma) h = 1/2 mv^2

(ma) x 89 = 0.5 x m x 53 x 53

a = (0.5 x 53 x 53)/89 = 15.8 m/s^2

Hence, (C) 15.8 m/s^2 is the correct ans.

(m/s is the unit of speed given in options, its should be m/s^2)