When a player thrown a ball, it reaches the other player in \(4 \mathrm{~s}\). If the height of each player…
When a player thrown a ball, it reaches the other player in \(4 \mathrm{~s}\). If the height of each player is \(1.8 \mathrm{~m}\), the maximum height attained by the ball above the ground is
\(19.4 \mathrm{~m}\)
\(20.4 \mathrm{~m}\)
\(21.4 \mathrm{~m}\)
\(22.4 \mathrm{~m}\)
Solution
Time taken to reach the ball from one player to another is equal to time of flight.
Hence, time of flight, \(T=4 \mathrm{~s}\)
\(\frac{2 u \sin \theta}{g}=4 \Rightarrow u \sin \theta=2 g\) ...(i)
Height attained by the ball above the ground \(=\) height of either player + maximum height
\(\begin{aligned}
& =1.8+\frac{u^2 \sin ^2 \theta}{2 g}=1.8+\frac{(u \sin \theta)^2}{2 g}=1.8+\frac{(2 g)^2}{2 g} \\
& =1.8+2 g=1.8+2 \times 9.8=1.8+19.6=21.4 \mathrm{~m}
\end{aligned}\)