Question:
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 $\mathrm{F}$ on the ball and moves horizontally a distance of $0.2 \mathrm{~m}$ while launching the ball, the value of $\mathrm{F}($ in $\mathrm{N})$ is $\left(\mathrm{g}=10 \mathrm{~ms}^{-2}\right)$__________.
Solution:
$\mathrm{W}_{\mathrm{F}}=\frac{1}{2} \mathrm{mv}^{2}=\mathrm{mgh}$
$\mathrm{F}(\mathrm{S})=\mathrm{mgh}$
$\mathrm{F}(0.2)=(0.15)(10)(20)$
$\mathrm{F}=150 \mathrm{~N}$
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