Question

A ball is thrown vertically upward from the top of a tower. Velocity at depth $h$ from the point of projection is twice of the velocity at height $h$ above the point of projection. Find the maximum height reached by the ball.

• A. $2h$
• B. $3h$
• C. $\left(\frac{5}{3}\right)h$
• D. $\left(\frac{4}{3}\right)h$

Answer 1

The correct option is C $\left(\frac{5}{3}\right)h$


Here, $|v_B|=2|v_A|$

By, applying 3rd equation of kinematics from A to B.
$u=v_A$
$v=v_B$
$v^2-u^2=2as$
$\Rightarrow v_B^2-v_A^2=2(-g)(-2h)4v_A^2-v_A^2=4gh3v_A^2=4gh{v}_A^2=\frac{4gh}{3}....\left(1\right)$

Again using 3rd equation of kinematics from A to C:
$v^2-u^2=2as$, here $v=0,u=v_A$
$0-v_A^2=2(-g)ss=\frac{v_A^2}{2g},from\left(1\right)s=\frac{4gh}{3\times 2g}$
$s=\frac{2h}{3}$

So maximum height reached by the ball $=s+h=\frac{5h}{3}$