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

$18.9 c m / s e c$

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$12.6 c m / s e c$

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$37.8 c m / s e c$

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None of these

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Solution

The correct option is C
$37.8 c m / s e c$

Step 1: Given Data:
Equation of motion, $s = u t - 6.3 t^{2}$
Stone reaches a maximum height in $t = 3 s e c$
Step 2: Formula used:
Velocity-displacement relation-
$v = \frac{d s}{d t}$
At maximum height, the particle's velocity at maximum height, $\frac{d s}{d t} = 0$
Step 3: Calculation of $u$
Given the equation of motion is- $s = u t - 6.3 t^{2}$
Using the formula,
$\frac{d (u t - 6.3 t^{2})}{d t} = 0$
$\begin{array}{rcl} u - 12.6 t & = & 0 \\ u & = & 12.6 t \\ = & 12.6 \times 3 \\ = & 37.8 c m / s \end{array}$
Hence, option C is the correct answer.

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