AIIMS 2019 Physics Projectile Motion MCQ Question

To determine the speed at which the cricketer throws the ball, we can use the projectile motion formula for maximum horizontal distance (range), given by $R = \frac{v^2 \sin(2\theta)}{g}$. For maximum range, the angle $\theta$ is 45°, which gives $\sin(90°) = 1$. Rearranging the formula, we find $v = \sqrt{Rg}$. Substituting $R = 100 \, \text{m}$ and $g \approx 9.81 \, \text{m/s}^2$, we get $v \approx \sqrt{100 \times 9.81} \approx 31.3 \, \text{m/s}$, rounding to 32 m/s, which corresponds to option C.

Options A (30 m/s), B (42 m/s), and D (35 m/s) are incorrect as they do not align with the calculated speed required to achieve the maximum range of 100 m under ideal conditions.