A Ball Of Mass 50g Is Dropped From A Height Of 20m. A Boy On The Ground Hits The Ball Vertically Upwards With A Bat With An Average Force Of 200 N, So That It Attains A Vertical Height Of 45 M. Find The Time For Which The Ball Remains In Contact With The Bat.

Sol:

Velocity of the ball just before hitting the bat = V_{1}

V_{1}^{2} = 0^{2} + 2gs

= 2 * 10 * 20 = 400

V_{1} = -20 m/sec

V_{1} is downwards

The height attained by the ball after being hit is -45 m

The speed just after being hit by the bat = V_{2}

= 0 = V_{2}^{2} – 2gs

= V_{2}^{2} = 2 * 10 * 45 = 900

V_{2}^{2} = 30 m/sec

V_{2}^{2} is upwards

Change in velocity = V_{2} – V_{1}

= 30 – (-20)

= 50 m/s.

Change in momentum = m V_{2} – m V_{1} = mΔV

= \(\frac{500}{1000} * 50 = 2.5 kd m/s\)

Force =\( \frac{\Delta p}{\Delta t} = 200 N\) \(\Delta t = \frac{2.5}{200} = \frac{1}{80} sec\)

Therefore, the time for which the ball remains in contact with the bat is\( \frac{1}{80} \)sec

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