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
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
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
Let the body after time $t/2$ be at $x$ from the top, then
$x = \frac{1}{2}g\frac{{{t^2}}}{4} = \frac{{g{t^2}}}{8}$…(i)
$h = \frac{1}{2}g{t^2}$…(ii)
Eliminate $t$ from (i) and (ii), we get $x = \frac{h}{4}$
$\therefore $ Height of the body from the ground $ = h - \frac{h}{4} = \frac{{3h}}{4}$
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