Question

A ball is thrown vertically upwards from the top of a tower. Velocity at a point ${}^{\prime }{h}^{\prime }m$ vertically below the point of projection is twice the downward velocity at a point ${}^{\prime }{h}^{\prime }m$ vertically above the point of projection. The maximum height reached by the ball above the top of the tower is:

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

Answer 1

Answer: (B)

$\frac{5h}{3}$

Let the velocity at height h above the tower be $v$ and kinetic energy at this point be $K$. Then the velocity at height h below the tower will be $2v$. Kinetic energy at this point will then be $4K$ since $K\alpha v^2$. Thus, by energy conservation,

$K+mgh=4K-mgh$

$K=\frac{2mgh}{3}$

Also, let the maximum height reached above tower be $H$. Then, again by conserving energy,

$K+mgh=\frac{2mgh}{3}+mgh=mgH$

Hence, $H=\frac{5h}{3}$