Question

When a player throws a ball, it reaches the other player in $4$ seconds. If the height of each player is $1.8\, m$, the maximum height attained by the ball above the ground is

• A. 19.4 m
• B. 20.4 m
• C. 21.4 m
• D. 22.4 m

Answer 1

Answer: (C)

Time taken to reach the ball from one player to another is equal to time of flight.
Hence, time of flight, $T=4 s$
$\frac{2 u \sin \theta}{g}=4$
$\Rightarrow u \sin \theta=2 g$... (i)
Height attained by the ball above the ground
$=$ height of either player + maximum height
$=1.8+\frac{u^{2} \sin ^{2} \theta}{2 g}=1.8+\frac{(u \sin \theta)^{2}}{2 g}$
$=1.8+\frac{(2 g)^{2}}{2 g}$
$=1.8+2 g=1.8+2 \times 9.8$
$=1.8+19.6=21.4\, m$