Question

A dynamite blast blows a heavy rock straight up with a launch velocity of $160$ $m/sec$. It reaches a height of $s=160t-16t^2$ after $t$ $sec$. The velocity of the rock when it is $256$ $m$ above the ground on the way up is

• A. $98$ $m/s$
• B. $96$ $m/s$
• C. $104$ $m/s$
• D. $48$ $m/s$

Answer 1

The correct option is B $96$ $m/s$

We know, $v=\frac{ds}{dt}=160-32t$.
We now find values of $t$ for which $s\left(t\right)=256$ So
$160t-16t^2=256$
$\Rightarrow 16(t^2-10t+16)=0$
$\Rightarrow (t-2)(t-8)=0$
$\Rightarrow t=2$, $t=8$.
So $v\left(2\right)=160-32.2=96$
$v\left(8\right)=160-256=-96$
So the velocity on the way up in $96$ m/s.