Question

A particle is projected at an angle of \( 30^\circ \) from horizontal at a speed of 60 m/s. The height traversed by the particle in the first second is \( h_0 \) and height traversed in the last second, before it reaches the maximum height, is \( h_1 \). The ratio \( \frac{h_0}{h_1} \) is __________. [Take \( g = 10 \, \text{m/s}^2 \)]

Hint:

For projectile motion, use the vertical component of velocity to calculate heights in different time intervals.

Answer 1

Correct Answer: 5

The vertical component of the velocity is: \[ 60 \sin 30^\circ = 30 \, \text{m/s} \] The height traversed in the first second \( S_1 \) is: \[ S_1 = 30 \times 1 - \frac{1}{2} \times 10 \times 1^2 = 25 \, \text{m} \] The height traversed in the last second \( S_3 \) is: \[ S_3 = 30 + \left( \frac{-10}{2} \right) \times (2 \times 3 - 1) = 5 \, \text{m} \] Thus, the ratio is: \[ \frac{S_1}{S_3} = \frac{25}{5} = 5 \]