Q 11-05-170JEE MainJEE Main 2020 (3 Sep, Shift 1)Medium
A cricket ball of mass $0.15\ \text{kg}$ is thrown vertically up by a bowling machine so that it rises to a maximum height of $20\ \text{m}$ after leaving the machine. If the part pushing the ball applies a constant force $F$ on the ball and moves horizontally a distance of $0.2\ \text{m}$ while launching the ball, the value of $F$ (in N) is ______ $(g = 10\ \text{m s}^{-2})$
Numerical value type. Enter your answer.
Answer: 150
Launch speed: $v^{2} = 2gh = 400$, so the kinetic energy given to the ball is $\tfrac12\times0.15\times400 = 30$ J.
This equals the work done by $F$ over $0.2$ m (neglecting the small change in potential energy during the push):
$$F\times0.2 = 30 \Rightarrow F = 150\ \text{N}$$
Solution by Sreeraj P, M.Sc Physics