## College Physics (4th Edition)

(a) The rocket's altitude when the engine fails is $25.0~km$ (b) The total time to reach maximum height is $152.0~s$ (c) The rocket travels up to a maximum height of $76.0~km$ (d) The velocity of the rocket just before it hits the ground is $1220~m/s$ downward.
(a) We can find the vertical displacement while the rocket accelerates upward: $\Delta y = \frac{1}{2}at^2$ $\Delta y = \frac{1}{2}(20.0~m/s^2)(50.0~s)^2$ $\Delta y = 25,000~m = 25.0~km$ The rocket's altitude when the engine fails is $25.0~km$ (b) We can find the velocity after the first $50.0~s$: $v_f = v_0+at$ $v_f = 0+(20.0~m/s^2)(50.0~s)$ $v_f = 1000~m/s$ We can find the time $t$ from the end of the upward acceleration period until maximum height. For this part, we can let $v_0 = 1000~m/s$: $v_f = v_0+at$ $t = \frac{v_f-v_0}{a}$ $t = \frac{0-1000~m/s}{-9.80~m/s^2}$ $t = 102.0~s$ The total time to reach maximum height is $152.0~s$ (c) We can find the displacement from the end of the upward acceleration period until the rocket reaches its maximum height. For this part, we can let $v_0 = 1000~m/s$: $v_f^2 = v_0^2+2a\Delta y$ $\Delta y = \frac{v_f^2-v_0^2}{2a}$ $\Delta y = \frac{0-(1000~m/s)^2}{(2)(-9.80~m/s^2)}$ $\Delta y = 51,020~m = 51.0~km$ The total displacement from the ground to the maximum height is $25.0~km+51.0~km$ which is $76.0~km$ The rocket travels up to a maximum height of $76.0~km$ (d) We can find the velocity just before the rocket hits the ground: $v_f^2=v_0^2+2a\Delta y$ $v_f=\sqrt{v_0^2+2a\Delta y}$ $v_f=\sqrt{0+(2)(-9.80~m/s^2)(-76,000~m)}$ $v_f = 1220~m/s$ The velocity of the rocket just before it hits the ground is $1220~m/s$ downward.