Question

A ball is thrown from the top of a tower in vertically upward direction. Velocity at a point $h \,m$ below the point of projection is twice of the velocity at a point $h \,m$ above the point of projection. Find the maximum height reached by the ball above the top of tower.

• A. 2h
• B. 3h
• C. (5/3)h
• D. (4/3)h

Answer 1

Answer: (C)

Maximum height reached, $H = \frac{u^2}{2g}$
Given $v_2 = 2v_1 $
$\Rightarrow \frac{v_1}{v_2} = \frac{1}{2} \,...(i)$
Motion from $A$ to $B$:
$v_1^2 = u^2 - 2 gh\, ... (ii)$
Motion from $A$ to $C$:
$v_2^2 = u^2 - 2g(-h)\, ...(iii)$
From $(ii)/(iii)$,
$\left(\frac{v_{1^{2}}}{v_{2}}\right)^{2} = \frac{u^{2} - 2gh}{u^{2} + 2gh} = \frac{\left(\frac{u^{2}}{2g}-h\right)}{\left(\frac{u^{2}}{2g} + h\right)} $
$\frac{1}{4 } = \frac{\left(H - h\right)}{\left(H + h\right)}$
$ \Rightarrow 4\left(H - h\right) = H + h$
$(H = \frac{u^2}{2g})$
$\Rightarrow 3H = 2h$
$\Rightarrow H = \frac{5}{3} h$