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(B) The total height of the tower is $H$. The time taken to reach the ground is $T$. Using the equation of motion $s = ut + \frac{1}{2}at^2$,where $u

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$\frac{15 H}{16}$

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(B) The total height of the tower is $H$. The time taken to reach the ground is $T$. Using the equation of motion $s = ut + \frac{1}{2}at^2$,where $u = 0$ and $a = g$:$H = \frac{1}{2} g T^2 \Rightarrow g T^2 = 2H \dots (i)$Now,let $x$ be the distance covered by the ball from the top in time $t = \frac{T}{4}$:$x = \frac{1}{2} g \left( \frac{T}{4} ight)^2 = \frac{1}{2} g \frac{T^2}{16} = \frac{g T^2}{32}$Substituting the value of $g T^2$ from equation $(i)$:$x = \frac{2H}{32} = \frac{H}{16}$The height of the ball from the ground is the total height minus the distance covered from the top:$	ext{Height} = H - x = H - \frac{H}{16} = \frac{15H}{16}$

$A$ ball is released from the top of a tower of height $H \ m$. It takes $T \ s$ to reach the ground. The height of the ball from the ground after $\frac{T}{4} \ s$ is

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