First slide

Rectilinear Motion

Question

A body is released from the top of a tower of height H m. After 2 s, it is stopped and then instantaneously released. What will be its height after next 2 s?

Moderate

A

$(H - 5) m$
B

$(H - 10) m$
C

$(H - 20) m$
D

$(H - 40) m$

Solution

Distance covered by the object in first 2 s
$h_{1} = \frac{1}{2} g t^{2} = \frac{1}{2} \times 10 \times 2^{2} = 20 m$
Similarly, distance covered by the object in next 2 s will also be 20 m, hence the required height = H-20-20= H-40 m

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