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

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
\(=\mathrm{H}-\frac{2 \mathrm{H}}{32}=\frac{15 \mathrm{H}}{16}\)