The cricketer can throw the ball to a maximum horizontal distance of 100 m. To do this the ball has to be thrown with a velocity v at an angle of elevation equal to 45 degrees.

The range of an object thrown at a velocity v with an angle of elevation 45 degrees is S = v^2/g

100 = v^2/g

=> v^2 = 100*g

In the same throw, the maximum height of the ball can be derived from the equation v^2 - u^2 = 2*g*h

It is is h = `(sqrt 980)^2/(2*9.8)` = `980/(2*9.8) = 50` m

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