Question

A ball is projected vertically up with a velocity of $20m{s}^{-1}$. Its velocity at the height of $15\text{m}$ is
(Take $g=10m{s}^{-2})$

• A. $10m{s}^{-1}$
• B. $15m{s}^{-1}$
• C. $-10m{s}^{-1}$
• D. $\text{Both a and c}$

Answer 1

The correct option is D $\text{Both a and c}$

Given, $u=20m{s}^{-1},s=15\text{m}$
We know that
$v^2-u^2=2as$
$v^2-(20{)}^2=-2\times 10\times 15$
$v^2=100\Rightarrow v=\pm 10m{s}^{-1}$
Hence, the ball will cross the height two times, first time when it goes up and second when it comes down. While going up its velocity at the height of $15\text{m}$ will be $10\text{m/s}$ and while coming down the velocity will be $-10\text{m/s}.$