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  4. The distance s in metres covered by a body, dropped from a tall cliff, as a function of time t in seconds is given by

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Q. The distance $s$ in metres covered by a body, dropped from a tall cliff, as a function of time $t$ in seconds is given by

Limits and Derivatives

Solution:

We know that,
$s=u t+\frac{1}{2} g t^2$, where
$s$ = distance, $u=$ initial velocity, $t=$ time,
$g$ - gravitational acceleration
Here, $ u=0, g=9.8 m / s ^2$
$ s=0 \times t+\frac{1}{2} \times 9.8 t^2 $
$ s=49 t^2 m \text { in } t s$