Question

1. a toy rocket starts from rest on the ground and then accelerates at 39.2 m/s² up for 5.0 s. (a) calculate the velocity of the rocket when the engines stop firing after 5.0 s. (b) calculate the maximum height of the rocket. (c) how long will it take the rocket to hit the ground from rest at the maximum height and initial rest?

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) $22.4\ s$