A ball of mass 400 gm is dropped from a height of 5m. A boy on the ground hits the ball vertically upwards with a bat with an average force of 100 newton so that it attains a vertical height of 20 m. The time for which the ball remains in contact with the bat is \(\left[ g = 10\, m/s^{2} \right]\)
Text Solution
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Velocity by which the ball hits the bat
\(v_{1} = \sqrt{2gh_{1}} = \sqrt{2 \times 10 \times 5}\) or \(\vec{v}_1 = +10 \, m/s = 10 \, m/s\)
velocity of rebound
\(v_{2} = \sqrt{2gh_{2}} = \sqrt{2 \times 10 \times 20} = 20 \, m/s\) or \(\vec{v}_2 = -20 \, m/s\)
\(F = m \frac{dv}{dt} = \frac{m(\vec{v}_2 - \vec{v}_1)}{dt} = \frac{0.4(-20 - 10)}{dt} = 100\,N\)
by solving dt = 0.12 sec
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