Question

The equation of motion of a stone thrown vertically upwards from the surface of a planet is given by $s=10t–3t^2$, and the units of $s$ and $t$ are cm and sec respectively. The stone will return to the surface of the planet after

• A. $\frac{10}{3}sec$
• B. $\frac{5}{3}sec$
• C. $\frac{20}{3}sec$
• D. $\frac{5}{6}sec$

Answer 1

The correct option is A

$\frac{10}{3}sec$

Step 1: Given Data:

The equation of motion,$s=10t–3t^2·····\left(1\right)$

Step 2: Formula used:

The relation between velocity and displacement is given by-

$v=\frac{ds}{dt}$

Step 3: Calculation of time:

Differentiate the equation $\left(1\right)$ with respect to time to obtain velocity,

$\begin{array}{rcl}\frac{ds}{dt}& =& \frac{d\left(10t-3t^2\right)}{dt}\\ v& =& 10-6t\end{array}$

But $v=0$ at maximum height. So,

$\begin{array}{rcl}0& =& 10-6t\\ 6t& =& 10\\ t& =& \frac{5}{3}\end{array}$

Therefore, the stone will return in $2\times \left(\frac{5}{3}\right)=\frac{10}{3}sec$ as it reached a maximum height in $\frac{5}{3}sec$ similarly it will take the same time to return to the surface of the planet.

Hence, option A is the correct answer.