Question

When a player throws a ball, it reaches the other player in $4$ seconds. If the height of each player is $1.8m$, 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, $T=4s$
$\frac{2u\sin \theta }{g}=4$
$\Rightarrow u\sin \theta =2g$... (i)
Height attained by the ball above the ground
$=$ height of either player + maximum height
$=1.8+\frac{u^2{\sin }^2\theta }{2g}=1.8+\frac{(u\sin \theta {)}^2}{2g}$
$=1.8+\frac{(2g{)}^2}{2g}$
$=1.8+2g=1.8+2\times 9.8$
$=1.8+19.6=21.4m$