Question

A Particle is projected vertically upward reaches 136 m height. What will be the maximum range for the particle projected with same speed ?

• A. 272 m
• B. 280 m
• C. 290 m
• D. 300 m

Answer 1

The Correct Option is A

To solve this problem, we need to analyze two scenarios involving projectile motion and vertical motion. The problem provides that a particle is projected vertically upward to reach a maximum height of 136 m, and we need to determine the maximum range when the same particle is projected horizontally with the same initial speed.

First, let's calculate the initial speed used by the particle to reach 136 m when projected vertically:

1. Using the kinematic equation for motion under gravity, where final velocity \( v = 0 \) at maximum height: v^2 = u^2 - 2gh
   • 0 = u^2 - 2 \times 9.8 \times 136
   • Rearrange to find initial speed \( u \): u^2 = 2 \times 9.8 \times 136
   • Calculate: u = \sqrt{2 \times 9.8 \times 136} = \sqrt{2665.6} \approx 51.63 \, \text{m/s}
2. Now, consider the particle projected horizontally to achieve maximum range. For maximum range, the angle of projection should be 45°:
3. Using the formula for range \( R \) of a projectile: R = \frac{u^2 \sin 2\theta}{g}
4. Here, \theta = 45^\circ:
5. Therefore: R = \frac{u^2 \sin 90^\circ}{g} since \sin 90^\circ = 1
6. Substitute the values:
   • R = \frac{(51.63)^2 \times 1}{9.8}
   • R = \frac{2665.6}{9.8} \approx 272 \, \text{m}

Thus, the maximum range for a particle projected with the same speed is 272 m. Hence, the correct option is:

• 272 m