## College Physics (4th Edition)

(a) The rocket travels up to a height of $54.85~m$ (b) The rocket returns to the ground 7.53 seconds after liftoff.
(a) We can find the vertical displacement while the rocket accelerates upward: $\Delta y = \frac{1}{2}at^2$ $\Delta y = \frac{1}{2}(17.5~m/s^2)(1.5~s)^2$ $\Delta y = 19.69~m$ We can find the velocity after the first $1.5~s$: $v_f = v_0+at$ $v_f = 0+(17.5~m/s^2)(1.5~s)$ $v_f = 26.25~m/s$ 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 $v_0 = 26.25~m/s$: $v_f^2 = v_0^2+2a\Delta y$ $\Delta y = \frac{v_f^2-v_0^2}{2a}$ $\Delta y = \frac{0-(26.25~m/s)^2}{(2)(-9.80~m/s^2)}$ $\Delta y = 35.16~m$ The total displacement from the ground to the maximum height is $19.69~m+35.16~m$ which is $54.85~m$ The rocket travels up to a height of $54.85~m$ (b) The rocket accelerates upward for a time of $t_1 = 1.5~s$ We can find the time $t_2$ from the end of the upward acceleration period until maximum height: $v_f = v_0+at_2$ $t_2 = \frac{v_f-v_0}{a}$ $t_2 = \frac{0-26.25~m/s}{-9.80~m/s^2}$ $t_2 = 2.68~s$ We can find the time $t_3$ for the rocket to fall from maximum height back down to the ground: $\Delta y = \frac{1}{2}at_3^2$ $t_3 = \sqrt{\frac{2\Delta y}{a}}$ $t_3 = \sqrt{\frac{(2)(-54.85~m)}{-9.80~m/s^2}}$ $t_3 = 3.35~s$ We can find the total time $t$: $t = t_1+t_2+t_3$ $t = 1.5~s+2.68~s+3.35~s$ $t = 7.53~s$ The rocket returns to the ground 7.53 seconds after liftoff.