Question

A ball A is a projected from the ground such that its horizontal range is maximum. Another ball left it is dropped from a height equal from a height equal to the maximum range of A?

Answer 1

In first ball,

$T=\frac{2usin\theta }{g}$

$T\left(A\right)=\frac{2usin{45}^o}{g}$

$T\left(A\right)=\frac{1.41u}{g}$

$R_{max}=\frac{u^2}{g}$

In second ball,

$s=ut+\frac{1}{2}{gt}^2$

$R_{max}=ut+\frac{1}{2}g{T_B}^2$

$T_B=\sqrt{\frac{{2R}_{max}}{g}}=\sqrt{\frac{2u^2}{g^2}}=\sqrt{\frac{2u}{g}}$

$\frac{T\left(A\right)}{T\left(B\right)}=\frac{1.41}{\sqrt{2}}=1$