7. a toy rocket starts from rest on the ground and then accelerates at 39.2 m/s² up for 5.0 s. (a) calculate…

Explanation:

Step1: Calculate velocity at engine - stop

We use the kinematic equation $v = v_0+at$. The initial velocity $v_0 = 0$ (starts from rest), acceleration $a = 39.2\ m/s^{2}$ and time $t = 5.0\ s$. $v=0 + 39.2\times5.0$ $v = 196\ m/s$

Step2: Calculate height during acceleration phase

We use the kinematic equation $y_1=v_0t+\frac{1}{2}at^{2}$. Since $v_0 = 0$, $a = 39.2\ m/s^{2}$ and $t = 5.0\ s$. $y_1=0\times t+\frac{1}{2}\times39.2\times5.0^{2}$ $y_1=\frac{1}{2}\times39.2\times25=490\ m$

Step3: Calculate height during free - fall phase

After the engines stop, the rocket continues to move up until its velocity becomes zero. We use the kinematic equation $v^{2}-v_{1}^{2}=- 2gy_2$, where $v = 0$ (at maximum height), $v_1 = 196\ m/s$ (velocity at engine - stop) and $g = 9.8\ m/s^{2}$. $0 - 196^{2}=-2\times9.8\times y_2$ $y_2=\frac{196^{2}}{2\times9.8}=1960\ m$ The maximum height $y_{max}=y_1 + y_2=490+1960 = 2450\ m$

Step4: Calculate time to fall from maximum height

We use the kinematic equation $y_{max}=\frac{1}{2}gt_2^{2}$, where $y_{max}=2450\ m$ and $g = 9.8\ m/s^{2}$. $t_2=\sqrt{\frac{2y_{max}}{g}}=\sqrt{\frac{2\times2450}{9.8}}=\sqrt{500}\approx22.4\ s$

Answer:

(a) $196\ m/s$ (b) $2450\ m$ (c) $22.4\ s$