Question

A ball is thrown from the top of a tower in vertically upward direction. Velocity at a point $hm$ below the point of projection is twice of the velocity at a point $hm$ 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 $B$:

$v_1^2=u^2-2gh...(ii)$

Motion from $A$ to $C$:

$v_2^2=u^2-2g(-h)...(iii)$

From $(ii)/(iii)$,

${\left(\frac{v_{1^2}}{v_2}\right)}^2=\frac{u^2-2gh}{u^2+2gh}=\frac{\left(\frac{u^2}{2g}-h\right)}{\left(\frac{u^2}{2g}+h\right)}$

$\frac{1}{4}=\frac{\left(H-h\right)}{\left(H+h\right)}$

$\Rightarrow 4\left(H-h\right)=H+h$

$(H=\frac{u^2}{2g})$

$\Rightarrow 3H=2h$

$\Rightarrow H=\frac{5}{3}h$