A cricket ball of mass $0.15 \mathrm{~kg}$ is thrown vertically up by a bowling machine so that it rises to a maximum height of 20 $\mathrm{m}$ after leaving the machine. If the part pushing the ball applies a constant force $F$ on the ball and moves
horizontally a distance of $0.2 \mathrm{~m}$ while launching the ball, the value of $F$ (in N) is $\left(g=10 \mathrm{~ms}^{-2}\right)$______
$(150.00)$
From work energy theorem, $W=F \cdot s=\Delta K E=\frac{1}{2} m v^{2}$
Here $V^{2}=2 g h$
$\therefore F \cdot s=F \times \frac{2}{10}=\frac{1}{2} \times \frac{15}{100} \times 2 \times 10 \times 20$
$\therefore F=150 \mathrm{~N}$
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