A cricketer hits a ball from the ground level with a velocity of $\left(10\sqrt{2}\right){\text{ms}}^{-1}$ at an angle of ${45}^{\circ }$ with the horizontal. What should be the minimum speed of the fielder standing [Along the direction in which the ball is hit] at a distance of $5\text{m}$ from the batsman so as to catch the ball at ground level. Assume the direction of motion of the ball and the fielder are the same. (Take $g=10m{s}^{-2})$

The correct option is B $7.5\text{m/s}$
Given $\theta ={45}^{\circ }$. Let $T$ be the time taken by the fielder to catch the ball.
Range $R=\frac{u^2\sin 2\theta }{g}=\frac{(10\sqrt{2}{)}^2\sin {90}^{\circ }}{10}=20\text{m}$
If the fielder needs to catch the ball then the time taken by him must be equal to the time of flight of the ball.
i.e $T=\frac{2u\sin \theta }{g}=\frac{2\times 10\sqrt{2}\times \frac{1}{\sqrt{2}}}{10}=2\text{s}$
Distance covered by the fielder to catch the ball $=R-5=20-5=15\text{m}$
Let $v$ be the minimum speed of the fielder to catch the ball.
Then, $15=v\times T$
$\Rightarrow v=\frac{15}{2}=7.5\text{m/s}$