## In This post you will get answer of Q.

A cricket ball thrown across a field is at heights $h_1$ and $h_2$ from the point of projection at times $t_1$ and $t_2$ respectively after the throw. The ball is caught by a fielder at the same height as that of projection. The time of flight of the ball in this journey is

Solution:

## $h_{1}=left(u,sin,thetaright)t_{1}-frac{1}{2}gt^{2}_{1} ,; ,h_{2}=left(u,sin,theta right)t_{2}-frac{1}{2}gt^{2}_{2}$

$Rightarrow frac{h_{1}+frac{1}{2}gt^{2}_{1}}{h_{2}+frac{1}{2}gt^{2}_{2}}=frac{t_{1}}{t_{2}} Rightarrow h_{1}t_{2}-h_{2}t_{1}=frac{g}{2}left(t_{1}t^{2}_{2}-t^{2}_{1}t_{2}right)$

$T=frac{2u,sin,theta}{g}=frac{2}{g}left[frac{h_{1}+frac{1}{2}gt^{2}_{1}}{t_{1}}right]=frac{2}{t_{1}} left[frac{h_{1}}{g}+frac{t^{2}_{1}}{2}right]=frac{h_{1}}{t_{1}}timesleft(frac{t_{1}t^{2}_{2}-t^{2}_{1}t_{2}}{h_{1}t_{2}-h_{2}t_{1}}right)+t_{1}=frac{h_{1}t^{2}_{2}-h_{2}t^{2}_{1}}{h_{1}t_{2}-h_{2}t_{1}}$