Question

A body is released from the top of a tower of height h. It takes $v=\frac{1}{2}b{t}^2+v_o$sec to reach the ground. Where will be the ball after time$\frac{t}{2}$sec.

• A. At h/2 from the ground
• B. At h/4 from the ground
• C. Depends upon mass and volume of the body
• D. At 3h/4 from the ground

Answer 1

The correct option is D At 3h/4 from the ground

let the body after time$\frac{t}{2}$be at x from the top,then

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

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

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

$\therefore$Height of the body from the ground $=h-\frac{h}{4}=\frac{3h}{4}$