Question

The equation of motion of a stone, thrown vertically upwards is , where the $s=ut–6.3t^2$units of $s$ and $t$ are cm and sec. if the stone reaches a maximum height in 3 sec, then $u=$

• A. $18.9cm/sec$
• B. $12.6cm/sec$
• C. $37.8cm/sec$
• D. None of these

Answer 1

The correct option is C

$37.8cm/sec$

Step 1: Given Data:

Equation of motion, $s=ut–6.3t^2$

Stone reaches a maximum height in $t=3sec$

Step 2: Formula used:

Velocity-displacement relation-

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

At maximum height, the particle's velocity at maximum height, $\frac{ds}{dt}=0$

Step 3: Calculation of $u$

Given the equation of motion is- $s=ut–6.3t^2$

Using the formula,

$\frac{d\left(ut-6.3t^2\right)}{dt}=0$

$\begin{array}{rcl}u–12.6t& =& 0\\ u& =& 12.6t\\ & =& 12.6\times 3\\ & =& 37.8cm/s\end{array}$

Hence, option C is the correct answer.