Question

When a ball is $h$ metre high from a point $O$, its velocity is $v$. When it is $hm$ below $O$, its velocity is $2v$. Find the maximum height from $O$ it will acquire.

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

Answer 1

Answer: (B)

$\frac{5h}{3}$

Using third equation of motion $v^2-u^2=2as$

For motion from Highest to lowest point $,{\left(2v\right)}^2-{\left(v\right)}^2=2g\left(h\right)or\frac{v^2}{2g}=\frac{2h}{3}$

We know Maximum height is given by $H=\frac{v^2}{2g}$
At TOP POINT velocity of ball is $v$, so maximum height it will attain $=\frac{v^2}{2g}=\frac{2h}{3}$
From point O,maximum height ball acquire $=h+\frac{2h}{3}=\frac{5h}{3}$