NCERT Solutions

We need to find the maximum horizontal range of a ball given its speed and a ceiling height constraint.

Concept: Projectile motion

Projectile motion describes the path of an object thrown into the air. The maximum height $H$ and horizontal range $R$ are given by:

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

where $u$ is the initial speed, $\theta$ is the launch angle, and $g$ is the acceleration due to gravity.

Given:

• Initial speed $u = 40\,\text{m}\cdot\text{s}^{-1}$
• Ceiling height $H = 25\,\text{m}$
• Acceleration due to gravity $g = 9.8\,\text{m}\cdot\text{s}^{-2}$

Solving:

1. Check the optimal angle:
   The maximum range for a projectile is achieved at $\theta = 45^{\circ}$. At this angle, the maximum height would be:

$$\begin{aligned} H_{45} &= \frac{40^2 \sin^2 45^{\circ}}{2 \cdot 9.8} \\ &= \frac{1600 \cdot 0.5}{19.6} \\ &\approx 40.8\,\text{m} \end{aligned}$$

Since $40.8\,\text{m} > 25\,\text{m}$, the ball would hit the ceiling. Thus, the maximum range is achieved when the ball just touches the ceiling ($H = 25\,\text{m}$).

2. Calculate the launch angle $\theta$:

$$\begin{aligned} H &= \frac{u^2 \sin^2 \theta}{2g} \\ 25 &= \frac{40^2 \sin^2 \theta}{2 \cdot 9.8} \\ \sin^2 \theta &= \frac{25 \cdot 19.6}{1600} \\ &= \frac{490}{1600} = \frac{49}{160} \\ \sin \theta &= \frac{7}{4\sqrt{10}} \end{aligned}$$

3. Calculate $\cos \theta$:

$$\begin{aligned} \cos \theta &= \sqrt{1 - \sin^2 \theta} \\ &= \sqrt{1 - \frac{49}{160}} \\ &= \sqrt{\frac{111}{160}} = \frac{\sqrt{111}}{4\sqrt{10}} \end{aligned}$$

4. Calculate the horizontal range $R$:

$$\begin{aligned} R &= \frac{u^2 \sin 2\theta}{g} \\ &= \frac{2u^2 \sin \theta \cos \theta}{g} \\ &= \frac{2 \cdot 1600 \cdot \frac{7}{4\sqrt{10}} \cdot \frac{\sqrt{111}}{4\sqrt{10}}}{9.8} \\ &= \frac{3200 \cdot \frac{7\sqrt{111}}{160}}{9.8} \\ &= \frac{140\sqrt{111}}{9.8} \\ &\approx 150.5\,\text{m} \end{aligned}$$

Answer

$$150.5\,\text{m}$$