Maximum horizontal distance, \\\\(R = 100 m\\\\) The cricketer will only be able to throw the ball to the maximum horizontal distance when the angle of projection is \\\\(45^ \\circ\\\\), i.e., \\\\(= 45 ^\\circ\\\\).The max horizontal range for a projection velocity v is given by the relation:\\\\(R_{max}=\\dfrac{u^2}{g}\\\\)\\\\(100= \\dfrac{u^2}{g}\\\\) The ball will achieve the maximum height when it is thrown vertically upward. For such motion, the final velocity v is zero at the maximum height H.Acceleration, \\\\(a = g\\\\)Using the third equation of motion:\\\\(v^2 - u^2= -2gH\\\\)\\\\(H = u^2 / 2g = 100 / 2 = 50 m\\\\)