Plzz quickly solve this question.....

$$\begin{gathered} Dear student \\ Solution to your b query is given below \\ Maximum horizontal dis\tan ce, R=100m \\ The cricketer will only be able to throw the ball to the maximum horizontal dis\tan ce \\ when the angle of projection is 45° \\ As R=\frac{u^2\sin \left(2\theta \right)}{g} \\ put \theta =45° \Rightarrow R_{max}=\frac{u^2\sin \left(2\times 45\right)}{g} \\ {R}_{max}=\frac{u^2\sin \left(90°\right)}{g} =\frac{u^2}{g} \\ \Rightarrow \frac{u^2}{g}=100m as given maximum horizontal dis\tan ce=100m \\ \Rightarrow u^2=100g \dots \dots i) \\ The ball will achieve the maximum height when it is thrown vertically upward. \\ In this case, the final velocity v is zero at the maximum height H. \\ Acceleration, a=-g \\ U\sin g the third equation of motion: \\ 2as=v^2-u^2 \\ \Rightarrow 2\times -g\times h_{max}=0^2-100g from (i) \\ \Rightarrow -2g{h}_{max}=-100g \\ \Rightarrow h_{max}=50m \\ So cricketer can throw the ball maximum to height equal to 50m \\ For remaining queries we request you to please initiate new chat session to \\ have rapid assis\tan ce from our experts \\ Regards \end{gathered}$$