A cricket ball of mass 0.15 kg is thrown vertically up by a bowling machine so that it rises to a ma

Question

A cricket ball of mass 0.15 kg is thrown vertically up by a bowling machine so that it rises to a maximum height of 20 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 m while launching the ball, the value of $F$ (in N ) is ( $g=10 \mathrm{~m} \mathrm{~s}^{-2}$ ) $\qquad$ .\

Professor PC Thomas Classes · Answer

(150) : Given : $m=0.15 \mathrm{~kg}$;\ $h=20 \mathrm{~m} ; x=0.2 \mathrm{~m} ; g=10 \mathrm{~m} / \mathrm{s}^{2}$\ From work-energy theorem,\ $F 	imes x=m g h$\ $\Rightarrow F 	imes 0.2=0.15 	imes 10 	imes 20$\ $	herefore F=150 \mathrm{~N}$\