Question

The height of ceiling in an auditorium is 30 m. A ball is thrown with a speed of \( 30 \, \text{m s}^{-1} \) from the entrance such that it just moves very near to the ceiling without touching it and then it reaches the ground at the end of the auditorium. Then the length of auditorium is [Acceleration due to gravity \( = 10 \, \text{m s}^{-2} \)]

• A. \( 60\sqrt{2} \, \text{m} \)
• B. \( 30\sqrt{2} \, \text{m} \)
• C. \( 70\sqrt{2} \, \text{m} \)
• D. \( 100\sqrt{2} \, \text{m} \)

Hint:

For projectile motion with initial speed \(u\) at angle \( \theta \) with horizontal: - Maximum Height: \( H_{max} = \frac{u^2 \sin^2\theta}{2g} \) - Horizontal Range: \( R = \frac{u^2 \sin(2\theta)}{g} = \frac{2u^2 \sin\theta \cos\theta}{g} \) Use \( \sin^2\theta + \cos^2\theta = 1 \) to find \( \cos\theta \) if \( \sin\theta \) is known (or vice-versa).

Answer 1

The Correct Option is A

Let the initial speed be \( u = 30 \, \text{m s}^{-1} \), and the angle of projection be \( \theta \).
The height of the ceiling is \( H_{ceiling} = 30 \, \text{m} \).
The ball just moves very near to the ceiling, so the maximum height of the projectile is \( H_{max} = 30 \, \text{m} \).
The formula for maximum height is \( H_{max} = \frac{u^2 \sin^2\theta}{2g} \).
Given \( g = 10 \, \text{m s}^{-2} \).
\[ 30 = \frac{(30)^2 \sin^2\theta}{2 \times 10} \] \[ 30 = \frac{900 \sin^2\theta}{20} \] \[ 30 = 45 \sin^2\theta \] \[ \sin^2\theta = \frac{30}{45} = \frac{2}{3} \] \[ \sin\theta = \sqrt{\frac{2}{3}} \] (since \( \theta \) is acute for projectile motion to reach maximum height).
Then \( \cos^2\theta = 1 - \sin^2\theta = 1 - \frac{2}{3} = \frac{1}{3} \).
So, \( \cos\theta = \sqrt{\frac{1}{3}} = \frac{1}{\sqrt{3}} \).
The length of the auditorium is the horizontal range \( R \) of the projectile.
The formula for range is \( R = \frac{u^2 \sin(2\theta)}{g} \).
\( \sin(2\theta) = 2\sin\theta\cos\theta = 2 \left(\sqrt{\frac{2}{3}}\right) \left(\frac{1}{\sqrt{3}}\right) = \frac{2\sqrt{2}}{3} \).
\[ R = \frac{(30)^2 \cdot \frac{2\sqrt{2}}{3}}{10} = \frac{900 \cdot \frac{2\sqrt{2}}{3}}{10} = \frac{300 \cdot 2\sqrt{2}}{10} = 30 \cdot 2\sqrt{2} = 60\sqrt{2} \, \text{m} \] The length of the auditorium is \( 60\sqrt{2} \, \text{m} \).
This matches option (1).