JEE Physics 2024 Question with Solution

Given: The ball is projected horizontally with speed $u$. Each step has height $0.1 \, \text{m}$ and width $0.1 \, \text{m}$. Find: the value of $x$ if $u = \sqrt{x} \, \text{m/s}$.

To just hit step $5$, the ball must reach the top edge of that step after crossing $4$ steps horizontally and falling $4$ steps vertically. Hence,

$$R = 4 \times 0.1 = 0.4 \, \text{m}$$

and

$$h = 4 \times 0.1 = 0.4 \, \text{m}$$

For horizontal motion,

$$R = ut$$

so

$$t = \frac{0.4}{u}$$

For vertical motion,

$$h = \frac{1}{2}gt^2$$

Thus,

$$0.4 = \frac{1}{2} \times 10 \times \left(\frac{0.4}{u}\right)^2$$ $$0.4 = 5 \cdot \frac{0.16}{u^2}$$ $$0.4 = \frac{0.8}{u^2}$$ $$u^2 = 2$$

Therefore,

$$u = \sqrt{2} \, \text{m/s}$$

Given $u = \sqrt{x} \, \text{m/s}$, we get

$$x = 2$$

Therefore, the required value is $2$.

The first provided approach computes the distances for step $5$ as $0.5 \, \text{m}$ horizontally and vertically, leading to $x = 2.5$. However, for the ball to just hit step 5, it must reach the beginning of the 5th step, not the far corner after crossing five full steps.

The second approach correctly uses horizontal and vertical distances as $0.4 \, \text{m}$ each, because the ball must cross only $4$ steps to arrive at step $5$. This gives

$$u^2 = 2$$

and hence

$$x = 2$$

Therefore, the defensible final answer is $2$.