AP EAPCET 06-Jul-2022 Shift 2 Paper

The motion of the ball is shown below. Range, $R\;=\;\frac{u^2\sin\;2\theta}{g}$$\;=\;\frac{900\sin\;(2\times 30)}{10}=\;\frac{900\times\sin\;60^{\circ}}{10}$$R\;=\;\frac{900\times\sqrt{3}}{10\times 2}=\;\frac{900\sqrt{3}}{20}=45\sqrt{3}\ \mathrm{m}$ Since, the IInd player is at $21\sqrt{3}\ \mathrm{~m}$ in the direction of kick and he is running to catch the ball, so this can be represented as shown Let the man runs with speed $v$ towards point $P$ to catch the ball just at point $P$ before it touches the ground in the same time when ball reaches the point $P$.The man will travel distance $=24\sqrt{3}\ \mathrm{m}$$\Rightarrow$ Distance $=$ speed $\times$ time$\Rightarrow\;\;2u\sqrt{3}=v\times t$......(i) Where, $t$ is the time of run as well as time of flight since both the ball and IInd man start running at same time hence $t=$ time of flightWe know that,Time of flight, $t=\;\frac{2u\sin\;\theta}{g}=\;\frac{2\times30\times\sin\;30^{\circ}}{10}$ $\;t=3\ \mathrm{s}$Put $t=3\ \mathrm{s}$ in Eq. (i)$\;2u\sqrt{3}\;=v\times3$$\Rightarrow v\;=8\sqrt{3}\ \mathrm{m}\ /\ \mathrm{s}$Option (c) is the correct answer.