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QJEEM20Ph6665: JEE Main 2020 Physics - Class 11 JEEM20Ph6665 Solution: A cricket ball of mass 0.15 kg is thrown vertically up by a…

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 ms–2) ____.

Step-by-Step Verified Solution

Initial velocity, v = 2gh\sqrt {2gh}

= 2×10×20\sqrt {2 \times 10 \times 20}

= 20 m/s

Now work done by the machine,

WF = Δ\Delta k

⇒\Rightarrow F.d = Δ\Delta k

⇒\Rightarrow F = Δkd{{\Delta k} \over d}

= 12×0.15×400−00.2{{{1 \over 2} \times 0.15 \times 400 - 0} \over {0.2}}

= 150 N

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