Chapter 6 - Dynamics I: Motion Along a Line - Exercises and Problems - Page 164: 50

a) Acceleration for the first $16s$ is $26.15 ms^{-2}$ b) velocity at $5100m $ above ground is $378.4ms^{-1}$

let $a$ be the acceleration of the rocket for $16s$ and $S_1$ be the distance(height) travelled in $16s$. From $16s$ to $20s$ acceleration of the rocket is $-g$ and $S_2$ be the distance travelled in $20s -16s =4s$. Let $v_1$ be the velocity after $16s$ and $v_2$ be the velocity after $20s$ Now $v_1 = u +at = 0 + 16a $ $v_1 =16a$ $S_1 =ut +\frac{1}{2}at^2$ $ S_1 = 0 + 128a $ $S_1= 128a$ $S_2 =v_1t -\frac{1}{2}gt^2$ $S_2 = 4v_1- 80$ a) we have $S_1 + S_2 =5100m$ , therefore $5100m = 128 a + 4 v_1 -80$ putting value of $v_1$ $5100 m = 128a +64a -80$ $ a= 26.15 ms^{-2}$ b) also we have $v_2 = v_1 -gt$ $v_2 = v_1 -40$ $v_2 = 16a -40$ $v_2 = (16* 26.15 -40) ms^{-1}$ $v_2 = 378.4ms^{-1}$