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)

28 Views|Posted 7 months ago
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7 months ago

First, find the initial velocity (v) of the ball as it leaves the machine, using the kinematic equation for maximum height:
v² = u² + 2as ⇒ 0 = v² - 2gh_max
v² = 2gh_max = 2 * 10 * 20 = 400
v = 20 m/s.
Now, apply the work-energy theorem to the ball while it is being pushed by the machine. The work done

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Physics Work, Energy and Power 2021

Physics Work, Energy and Power 2021

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