JEE Physics 2023 Question with Solution

Given: Height of cliff = $10 \, \text{m}$, horizontal speed = $5 \, \text{m/s}$, and $g = 10 \, \text{m/s}^2$.

Find: Speed with which the stone hits the ground.

The solution states that the vertical component of velocity on reaching the ground is

$$v_y = \sqrt{2gh} = \sqrt{200}$$

The horizontal component remains unchanged:

$$v_x = 5 \, \text{m/s}$$

So the net speed is

$$v_{\text{net}} = \sqrt{v_x^2 + v_y^2} = \sqrt{25 + 200} = \sqrt{225}$$

Therefore, the speed of the stone when it hits the ground is $15 \, \text{m/s}$.

The solution explicitly concludes The Correct Option is A and also writes 15. Since $15$ appears as option B in the listed options, there is a discrepancy between the option label and the option value. Using the worked value from the solution, the defensible answer is A as declared by the solution, while the computed value is $15 \, \text{m/s}$.

Given: The stone is projected horizontally with speed $5 \, \text{m/s}$ from a height of $10 \, \text{m}$.

Find: The resultant speed at impact.

For vertical motion, initial vertical velocity is zero. Using the kinematic relation shown in the solution:

$$v_y = \sqrt{2gh}$$

Substituting:

$$v_y = \sqrt{2 \times 10 \times 10} = \sqrt{200}$$

For horizontal motion, there is no acceleration, so

$$v_x = 5 \, \text{m/s}$$

Now combine the perpendicular velocity components:

$$v = \sqrt{v_x^2 + v_y^2}$$ $$v = \sqrt{5^2 + \left(\sqrt{200}\right)^2}$$ $$v = \sqrt{25 + 200} = \sqrt{225} = 15 \, \text{m/s}$$

Therefore, the impact speed is $15 \, \text{m/s}$.