Question

A ball is thrown vertically upward. It has a speed of $10m/s$ when it has reached one-half of its maximum height. How high does the ball rise? (Take, $g=10m/s^2)$

• A. 15m
• B. 10m
• C. 20m
• D. 5m

Answer 1

Answer: (B)

The problem can be solved using third equation of motion at $A$ and $O$.

Let maximum height attained by the ball be $H$.
Third equation of motion gives $v^2=u^2-2gh$
At $A,(10{)}^2=u^2-2\times 10\times \frac{H}{2}$
$\Rightarrow u^2=100+10H...(i)$
At $O,(0{)}^2=u^2-2\times 10\times H$
$\Rightarrow u^2=20H...(ii)$
Thus, from Eqs. (i) and (ii),
we get $20H=100+10H$
$\Rightarrow 10H=100$
$\therefore H=10m$