Question

A fielder on the ground throws a ball at an angle of ${15}^{\circ }$ to the horizontal with velocity $V_A$ at the wicket to dismiss the batsman. Had he thrown the ball at ${45}^{\circ }$, with a speed $V_B$ to hit the wicket, then $\frac{V_B}{V_A}$is

• A. $\frac{1}{\sqrt{3}}$
• B. $\sqrt{2}$
• C. $\sqrt{3}$
• D. $\frac{1}{\sqrt{2}}$

Answer 1

Answer: (B)

$\sqrt{2}$

To dismiss the batman the fielder has to make sure that the ball hits the wicket, so in both the cases the range of the ball needs to be the same.

Let $R_A$ be the range when ball is thrown with velocity $V_A$ at ${15}^o$

Let $R_B$ be the range when ball is thrown with velocity $V_B$ at ${45}^o$

$⟹$$R_A=R_B$

We know that for a ball thrown with velocity $u$ at an angle $\theta$, the range, $R$ is given by,

$R=\frac{u^{2}Sin2\theta }{g}$

$⟹$ $\frac{V_A^{2}Sin(2\ast {15}^o)}{g}$ = $\frac{V_B^{2}Sin(2\ast {45}^o)}{g}$

$⟹$$\frac{V_A^2}{2}=\frac{V_B^2}{1}$

$⟹$ $\frac{V_B}{V_A}=\sqrt{2}$