Question

a ball is hit with 20 m/s velocity creating an angle of 30 degree.it is dropped after some time.6 s after dropping, one fielder took the ball and threw it .at that moment a batsman achieves 1 run and started running for the 2nd run.after 3 s from throwing time , the ball hits the stump.to complete 1 run a batsman need minimum 6 s. will the batsman got run out ??

Answer 1

Answer on Question#50527 - Physics - Mechanics - Kinematics - Dynamics

A ball is hit with $v_0 = 20\,\mathrm{m/s}$ velocity creating an angle of $\varphi = 30{}^\circ$. It is dropped after some time. 6 s after dropping, one fielder took the ball and threw it. At that moment a batsman achieves 1 run and started running for the $2^{\text{nd}}$ run. After 3 s from throwing time, the ball hits the stump. To complete 1 run a batsman need minimum 6 s. Will the batsman got run out?

Solution:

After the ball was hit, it (the ball) was flying for some time $t$ which defines as follows

$$t = \frac{2 v_0 \cdot \sin \varphi}{g} = \frac{2 \cdot 20 \frac{\mathrm{m}}{\mathrm{s}} \cdot 0.5}{10 \frac{\mathrm{m}}{\mathrm{s}^2}} = 2\,\mathrm{s}$$

The total time which passed from when the ball was hit and till it hit the stump is

$$t_{\text{total}} = 2\,\mathrm{s} + 6\,\mathrm{s} + 3\,\mathrm{s} = 11\,\mathrm{s}$$

The batsman needs 12 s to achieve $2^{\text{nd}}$ run. Since it's larger than 11 s, he won't got run out.

Answer: won't got run out.

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