A ball is released from the top of a tower of height H m. It takes T second to reach the ground. The height…
A ball is released from the top of a tower of height H m. It takes T second to reach the ground. The height of the ball from the ground after $\frac{T}{4}$ second is
$\frac{13 \mathrm{H}}{14}$
$\frac{15 \mathrm{H}}{16}$
$\frac{11 \mathrm{H}}{12}$
$\frac{9 \mathrm{H}}{10}$
Solution
Let the body be at $x$ from the top after $\frac{t}{4} \mathrm{~s}$.
$\mathrm{x}=\mathrm{ut}+\frac{1}{2} a \mathrm{t}^2=\frac{1}{2} a \mathrm{t}^2$
$\ldots(\because u=0)$
$\therefore \quad x=\frac{1}{2} g\left(\frac{\mathrm{t}}{4}\right)^2=\frac{g t^2}{32} \Rightarrow 32 \mathrm{x}=\mathrm{gt}^2...(i)$
$\mathrm{H}=\frac{1}{2} \mathrm{gt}^2 \Rightarrow 2 \mathrm{H}=\mathrm{gt}^2...(ii)$ From (i) and (ii), we get
$2 \mathrm{H}=32 \mathrm{x} \Rightarrow \mathrm{x}=\frac{2 \mathrm{H}}{32}$
$\therefore \quad$ Height of the body from the ground
$=\mathrm{H}-\frac{2 \mathrm{H}}{32}=\frac{15 \mathrm{H}}{16}$
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