There is a hypothetical cricket match going on in kolkata between India vs Australia.
Presently there is a spinner of height 2 metre bowling first ball of the over to a right hander.He is bowling a leg stump line, his ball deceives the batsmen and he gets bowled.
###### Image credit: Wikipedia John Sutton

In the xyz,co-ordinate space consider the pitch to be the region \(0 \le x \le 20, -2 \le y \le 2,z=0\)

Consider stumps to be the lines

\(0 \le z \le 0.5,x=20,y=0.125\)

\(0 \le z \le 0.5,x=20,y=0\)

\(0 \le z \le 0.5,x=20,y=-0.125 \)

The ball is released from the point \((0,-0.125,2)\) with a velocity \(30 \hat{i}\) m/s, also given an angular velocity of \(-\omega \hat{i}, \omega \ge 0\).

The question is simple, if the maximum value of \(\omega\), for which the batsmen can get bowled is \(=\frac { a(b+c\sqrt { b } ) }{ d }\) rad/s where \(a,b,c,d\) are integers, \(a,d\) are co-prime,\(b\) is not divisible by any square prime number then find \(a+b+c+d\).

**Details and Assumptions**

1) Assume ball a solid sphere of radius \(0.01 m, g= 10 m{s}^{-2}\).Consider the pitch to be sufficiently rough and the collision to be elastic.

2) All the co-ordinates are given in metres.

3) If in the trajectory of the ball it is hitting the stumps, consider the batsmen bowled.

4) Consider it as a point mass while calculating it's trajectory

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