Question

A ball projected horizontally with speed v from top of a tower of height h reaches the ground at distance R from the foot of the tower .Another ball projected horizontally from the top of atower of height 2h ,reaches the ground at a distance 2R from the foot of the tower what is the initial speed of the second ball

Answer 1

        $\text{\huge{\bf{Answer:}}}$

        $\text{Case I:}$

        $\text{Initial velocity} = v$

        $\text{Height} = h$

        $\text{Range} = R$

        $\text{We know that time taken to reach the ground for horizontal projectile is}$

        $T = \sqrt{\dfrac{2h}{g}}$

$\implies \; T_1 = \sqrt{\dfrac{2h}{g}}$

        $\text{Using Distance} =\text{Speed} \times \text{Time}$

$\implies \; R = vT_1$

$\implies \; R = v\sqrt{\dfrac{2h}{g}} \text{ ------ (i)}$

        $\text{Case II:}$

        $\text{Height} = 2h$

        $\text{Range} = 2R$

$\implies \; T_2 = \sqrt{\dfrac{4h}{g}}$

$\implies \; 2R = VT_2$

$\implies 2R = V\sqrt{\dfrac{4h}{g}} \text{ ------ (ii)}$

        $\text{Diving (i) by (ii)}$

$\implies \; \dfrac{R}{2R} = \dfrac{v\sqrt{\dfrac{2h}{g}}}{V\sqrt{\dfrac{4h}{g}}}$

$\implies \; \dfrac{1}{2} = \dfrac{v}{V\sqrt2}$

$\implies \; \dfrac{1}{\sqrt2} = \dfrac{v}{V}$

$\implies \; \boxed{V = \sqrt2v}$