JEE Physics 2025 Question with Solution

Given: The particle is projected with speed $60 \, \text{m/s}$ at angle $30^\circ$. Also, $g = 10 \, \text{m/s}^2$.

Find: The ratio $\frac{h_0}{h_1}$, where $h_0$ is the height traversed in the first second and $h_1$ is the height traversed in the last second before reaching maximum height.

The vertical component of velocity is

$$60 \sin 30^\circ = 30 \, \text{m/s}$$

So, $u_y = 30 \, \text{m/s}$.

The height traversed in the first second is

$$S_1 = 30 \times 1 - \frac{1}{2} \times 10 \times 1^2 = 25 \, \text{m}$$

Hence, $h_0 = 25 \, \text{m}$.

Time to reach maximum height is

$$t_{\text{max}} = \frac{u_y}{g} = \frac{30}{10} = 3 \, \text{s}$$

Therefore, the last second before maximum height is from $t = 2 \, \text{s}$ to $t = 3 \, \text{s}$.

Using $y = u_y t - \frac{1}{2}gt^2$,

$$y(3) = 30(3) - \frac{1}{2}(10)(3)^2 = 90 - 45 = 45 \, \text{m}$$ $$y(2) = 30(2) - \frac{1}{2}(10)(2)^2 = 60 - 20 = 40 \, \text{m}$$

So,

$$h_1 = y(3) - y(2) = 45 - 40 = 5 \, \text{m}$$

Thus,

$$\frac{h_0}{h_1} = \frac{25}{5} = 5$$

Therefore, the required ratio is $5$.

Given: $u_y = 30 \, \text{m/s}$ and $g = 10 \, \text{m/s}^2$.

Find: The ratio $\frac{h_0}{h_1}$.

For vertical motion, displacement in the $n$th second is

$$S_n = u_y + \frac{a}{2}(2n-1)$$

Here, $a = -10 \, \text{m/s}^2$.

For the first second,

$$S_1 = 30 + \frac{-10}{2}(2 \times 1 - 1) = 30 - 5 = 25 \, \text{m}$$

So, $h_0 = 25 \, \text{m}$.

The particle reaches maximum height in

$$\frac{30}{10} = 3 \, \text{s}$$

Hence, the last second before maximum height is the third second.

For the third second,

$$S_3 = 30 + \frac{-10}{2}(2 \times 3 - 1) = 30 - 25 = 5 \, \text{m}$$

So, $h_1 = 5 \, \text{m}$.

Therefore,

$$\frac{h_0}{h_1} = \frac{25}{5} = 5$$

Therefore, the required ratio is $5$.