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 s. Options: A . &nbsp At h 2 from the ground B . &nbsp At h 4 from the ground C . &nbsp Depends upon mass and volume of the body D . &nbsp At 3 h 4 from the ground

Answer: Option D : D Let the body after time t 2 fall a distance x from the top. Since it starts from rest, we have x = 1 2 g t 2 4 = g t 2 8 .............(i) It reaches the ground in t seconds, we have h = 1 2 g t 2 ..............(ii) Eliminate t from (i) and (ii), we get x = h 4 ∴ Height of the body from the ground = h − h 4 = 3 h 4

A body is released from the top of a tower of height h. It takes t sec to

Discussion Forum : Motion In One Dimension

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 $\frac{t}{2}$ s.

Options:

A . At

\frac{h}{2}

from the ground

B . At

\frac{h}{4}

from the ground

C . Depends upon mass and volume of the body

D . At

\frac{3 h}{4}

from the ground

Answer: Option D
:
D

Let the body after time $\frac{t}{2}$ fall a distance x from the top. Since it starts from rest, we have

$x = \frac{1}{2} g \frac{t^{2}}{4} = \frac{g t^{2}}{8}$ .............(i)

It reaches the ground in t seconds, we have

$h = \frac{1}{2} g t^{2}$ ..............(ii)

Eliminate t from (i) and (ii), we get x = $\frac{h}{4}$

$∴$ Height of the body from the ground

= $h - \frac{h}{4} = \frac{3 h}{4}$

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