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: \[ \frac{S_1}{S_3} = \frac{25}{5} = 5 \]

Answer 2

Step 2 — Time to reach maximum height: 
\[ t_{\text{max}} = \frac{u_y}{g} = \frac{30}{10} = 3\ \text{s}. \] Step 3 — Height at any time \(t\):
\[ y = u_y t - \frac{1}{2} g t^2. \] Step 4 — Height traversed in the first second:
At \(t=1\): \[ h_0 = y(1) - y(0) = (30 \times 1 - \frac{1}{2} \times 10 \times 1^2) = 30 - 5 = 25\ \text{m}. \] Step 5 — Height traversed in the last second (i.e., between \(t=2\) and \(t=3\)):
\[ h_1 = y(3) - y(2). \] Now, \[ y(3) = 30(3) - \frac{1}{2}(10)(3)^2 = 90 - 45 = 45\ \text{m}, \] \[ y(2) = 30(2) - \frac{1}{2}(10)(2)^2 = 60 - 20 = 40\ \text{m}. \] Thus, \[ h_1 = 45 - 40 = 5\ \text{m}. \] Step 6 — Ratio of the heights:
\[ \frac{h_0}{h_1} = \frac{25}{5} = 5. \] Final Answer: \[ \boxed{\dfrac{h_0}{h_1} = 5.} \]