A ball is dropped from the top of a 100; m hi | vedclass

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(B) Let the total time taken to reach the ground be $T$ seconds.Using the equation of motion $s = ut + \frac{1}{2}at^2$,where $u = 0$ (dropped from re

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$8$

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(B) Let the total time taken to reach the ground be $T$ seconds.Using the equation of motion $s = ut + \frac{1}{2}at^2$,where $u = 0$ (dropped from rest):For the total distance $100\; m$,we have $100 = \frac{1}{2}aT^2 \implies T = \sqrt{\frac{200}{a}}$.In the last $\frac{1}{2}\; s$,the ball covers $19\; m$. This means in time $(T - 0.5)\; s$,the ball covers $(100 - 19) = 81\; m$.So,$81 = \frac{1}{2}a(T - 0.5)^2 \implies T - 0.5 = \sqrt{\frac{162}{a}}$.Substituting $T = \sqrt{\frac{200}{a}}$ into the equation:$\sqrt{\frac{200}{a}} - 0.5 = \sqrt{\frac{162}{a}}$$\frac{10\sqrt{2}}{\sqrt{a}} - \frac{9\sqrt{2}}{\sqrt{a}} = 0.5$$\frac{\sqrt{2}}{\sqrt{a}} = 0.5$$\sqrt{\frac{2}{a}} = \frac{1}{2}$Squaring both sides: $\frac{2}{a} = \frac{1}{4} \implies a = 8\; m/s^2$.

$A$ ball is dropped from the top of a $100\; m$ high tower on a planet. In the last $\frac{1}{2}\; s$ before hitting the ground,it covers a distance of $19\; m$. Acceleration due to gravity (in $m/s^2$) near the surface on that planet is:

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