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JEE Mains · Physics · STD 11 - 5. work,energy,power and collision

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  \(\left(g=10\, m s^{-2}\right)\)

  1. A \(200\)
  2. B \(150\)
  3. C \(275\)
  4. D \(325\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(150\)

Step-by-step Solution

Detailed explanation

\(W _{ F }=\frac{1}{2} mv ^{2}= mgh\) \(F ( S )= mgh\) \(F (0.2)=(0.15)(10)(20)\) \(F =150 N\)
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