Question

A dynamite blast blows a heavy rock straight up with a launch velocity of 160 m/sec. It reaches a height of $s=160t-16{{t}^{2}}$ 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

Hint: Firstly, we have to find the velocity by differentiating $s=160t-16{{t}^{2}}$ with respect to t. Then, find the value of t when $s=256m$ by substituting it in the equation $s=160t-16{{t}^{2}}$ . Substitute these values of t in the velocity equation and simplify.

Complete step by step answer:
We are given that the velocity with which the rock is launched is 160 m/sec. We are given that the rock reaches the height $s=160t-16{{t}^{2}}$ . This means that $s=160t-16{{t}^{2}}$ is the distance.
We know that velocity is the derivative of distance with respect to time.
$\Rightarrow v=\dfrac{ds}{dt}$
We know that $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$ .
$\Rightarrow v\left( t \right)=160-32t...\left( i \right)$
We have to find the velocity when $s=256\text{ m}$ . Let us find t from $s=160t-16{{t}^{2}}$ by substituting this value of s.
$\begin{align}
  & \Rightarrow 256=160t-16{{t}^{2}} \\
 & \Rightarrow 16{{t}^{2}}-160t+256=0 \\
 & \Rightarrow 16\left( {{t}^{2}}-10t+16 \right)=0 \\
 & \Rightarrow {{t}^{2}}-10t+16=0 \\
\end{align}$
Let us factorize the above polynomial by splitting the middle term.
$\begin{align}
  & \Rightarrow {{t}^{2}}-2t-8t+16=0 \\
 & \Rightarrow t\left( t-2 \right)-8\left( t-2 \right)=0 \\
 & \Rightarrow \left( t-8 \right)\left( t-2 \right)=0 \\
 & \Rightarrow \left( t-8 \right)=0,\left( t-2 \right)=0 \\
 & \Rightarrow t=8\text{ sec },2\text{ sec} \\
\end{align}$
Now, let us find the value of $v\left( t \right)$ at $t=8\text{ and }t=2$ .
$\begin{align}
  & \Rightarrow v\left( 8 \right)=160-32\times 8 \\
 & \Rightarrow v\left( 8 \right)=160-256 \\
 & \Rightarrow v\left( 8 \right)=-96\text{ m/s} \\
\end{align}$
Now, we have to find $v\left( t \right)$ at $t=2$ .
$\begin{align}
  & \Rightarrow v\left( 2 \right)=160-32\times 2 \\
 & \Rightarrow v\left( 2 \right)=160-64 \\
 & \Rightarrow v\left( 2 \right)=96\text{ m/s} \\
\end{align}$
Therefore, the velocity of the rock when it is 256 m above the ground on the way up is 96 m/s.

So, the correct answer is “Option b”.

Note: Students must note that derivative of distance is velocity and derivative of velocity is acceleration. Acceleration is also the second order derivative of distance.
$\Rightarrow a=\dfrac{dv}{dt}=\dfrac{{{d}^{2}}s}{d{{t}^{2}}}$
Students must be thorough with the formulas of derivatives.