A tennis ball is thrown in the vertically upward direction and the bal

A tennis ball is thrown in the vertically upward direction and the ball attains a maximum height of 20 m. The ball was thrown approximately with an upward velocity of

8 m/s
12 m/s
16 m/s
20 m/s
This question was previously asked in
UPSC NDA-2 – 2021
The correct option is D. Using the kinematic equation $v^2 = u^2 + 2as$, where $v$ is the final velocity (0 m/s at max height), $u$ is the initial velocity, $a$ is the acceleration due to gravity (-g), and $s$ is the height (20 m).
At the maximum height of a vertically thrown object, its instantaneous velocity is zero. The acceleration acting on the object throughout its flight (ignoring air resistance) is the constant acceleration due to gravity, acting downwards. Taking the upward direction as positive, $a = -g$. Using $g \approx 10 \, \text{m/s}^2$ for approximation: $0^2 = u^2 + 2(-10)(20) \implies 0 = u^2 – 400 \implies u^2 = 400 \implies u = 20 \, \text{m/s}$. Using $g \approx 9.8 \, \text{m/s}^2$: $0^2 = u^2 + 2(-9.8)(20) \implies 0 = u^2 – 392 \implies u^2 = 392 \implies u \approx 19.8 \, \text{m/s}$. Both approximations are closest to 20 m/s among the given options.
This is a standard problem in projectile motion under constant acceleration (gravity). The time taken to reach the maximum height can be found using $v = u + at$. The total time of flight is twice the time to reach the maximum height (assuming it lands at the same level).