Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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?

At h/2 from the ground

At h/4 from the ground

Depends upon mass and volume of the body

At 3h/4 from the ground

Total Distance (Height h): For a body released from rest ( $u=0$ ), the distance covered in time $t$ is given by $h = \frac{1}{2}gt^2$ .
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 $h' = \frac{h}{4}$ .
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 = $h - h' = h - \frac{h}{4} = \frac{3h}{4}$ .
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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