Question

A cricketer hits a ball at $45^{\circ}$ with velocity $40\ \text{m s}^{-1}$ and it falls at $160\ \text{m}$. If he hits at the same angle with $50\ \text{m s}^{-1}$, the distance will be:

• A. 480 m
• B. 180 m
• C. 280 m
• D. 300 m
• E. 250 m

Hint:

For a fixed launch angle, doubling the velocity quadruples the range.

Answer 1

Answer: (E)

Step 1: Concept
Range of a projectile $R = \frac{u^2 \sin 2\theta}{g}$.

Step 2: Analysis
Since $\theta$ and $g$ are constant, $R \propto u^2$. $\frac{R_2}{R_1} = (\frac{u_2}{u_1})^2 = (\frac{50}{40})^2 = \frac{25}{16}$.

Step 3: Calculation
$R_2 = \frac{25}{16} \times 160 = 25 \times 10 = 250\ m$. Final Answer: (E)