Class 11 NCERT PHYSICS • Chapter 3: Motion in a Plane • Question 13

QUESTION 13

A cricketer can throw a ball to a maximum horizontal distance of 100 m. How much high above the ground can the cricketer throw the same ball ?

SOLUTION

We need to determine the maximum vertical height the cricketer can achieve given their maximum horizontal throw distance.

Concept: Projectile motion

The projectile motion equations for range ( $R$ ) and maximum height ( $H$ ) are used. For a given initial velocity $u$ , the maximum horizontal distance occurs at a launch angle of $\theta = 45^{\circ}$ .

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

For vertical motion, the maximum height reached when thrown straight up ( $\theta = 90^{\circ}$ ) is:

H_{\text{max}} = \frac{u^2}{2g}

Given:

Maximum horizontal distance, $R_{\text{max}} = 100\,\text{m}$
Acceleration due to gravity, $g \approx 9.8\,\text{m}\cdot\text{s}^{-2}$ (constant)

Solving:

Relate the maximum horizontal distance to the initial velocity

\begin{aligned} R_{\text{max}} &= \frac{u^2}{g} \\ 100\,\text{m} &= \frac{u^2}{g} \\ u^2 &= 100g \end{aligned}

Calculate the maximum vertical height when the ball is thrown straight upwards

When thrown vertically, the launch angle $\theta = 90^{\circ}$ .
Using the kinematic equation $v^2 = u^2 - 2gH$ , at the peak $v = 0$ :

\begin{aligned} H_{\text{max}} &= \frac{u^2}{2g} \\ &= \frac{100g}{2g} \\ &= \frac{100}{2} \\ &= 50\,\text{m} \end{aligned}

Answer:

50\,\text{m}