A body is released from the top of a tower of height $h$. It takes $t$ sec to reach the ground. Where will be

Question

A body is released from the top of a tower of height $h$. It takes $t$ sec to reach the ground. Where will be the ball after time $t/2$ sec?

Accepted Answer

1.  **Total Distance (Height h):** For a body released from rest ($u=0$), the distance covered in time $t$ is given by $h = \frac{1}{2}gt^2$ .
2.  **Distance covered in t/2:** Let $h'$ be the distance fallen from the top in time $t/2$.
    $h' = \frac{1}{2}g(t/2)^2 = \frac{1}{2}g \frac{t^2}{4} = \frac{1}{4} \left( \frac{1}{2}gt^2 \right)$.
    Since $h = \frac{1}{2}gt^2$, substituting this gives $h' = \frac{h}{4}$.
3.  **Position from Ground:** The distance calculated ($h'$) is measured from the release point (top). The height from the ground is total height minus distance fallen.
    Height from ground = $h - h' = h - \frac{h}{4} = \frac{3h}{4}$.
4.  **Alternative Method (Galileo's Law):** For equal time intervals (0 to t/2 and t/2 to t), distances traveled are in the ratio 1:3. Total distance (4 parts) corresponds to $h$. Distance fallen in first interval is 1 part ($h/4$). Remaining distance from ground is 3 parts ($3h/4$) .

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