JEE Physics 2024 Question with Solution

Given: The body falls under gravity from rest. The distance between A and B is $80 \, \text{m}$, the time taken is $2 \, \text{s}$, and $g = 10 \, \text{m/s}^2$.

Find: The distance of point A from the starting point.

For motion from A to B, use

$$s = ut + \frac{1}{2}at^2$$

Taking downward direction as positive,

$$80 = v_1(2) + \frac{1}{2}(10)(2^2)$$ $$80 = 2v_1 + 20$$ $$2v_1 = 60 \implies v_1 = 30 \, \text{m/s}$$

Now for motion from the starting point to A, the body starts from rest, so

$$v_1^2 = u^2 + 2gS$$ $$30^2 = 0 + 2(10)S$$ $$900 = 20S$$ $$S = 45 \, \text{m}$$

Therefore, the distance of point A from the starting point is $45 \, \text{m}$.

Given: Distance between A and B is $80 \, \text{m}$, time from A to B is $2 \, \text{s}$, and acceleration due to gravity is $10 \, \text{m/s}^2$.

Find: Distance from the starting point to A.

Let the velocity at point A be $v_1$.

From A to B,

$$80 = v_1 t + \frac{1}{2}gt^2$$

Substitute $t = 2 \, \text{s}$:

$$80 = 2v_1 + 20$$ $$2v_1 = 60$$ $$v_1 = 30 \, \text{m/s}$$

Now from start to A, initial velocity is zero. Use

$$v_1^2 = 2gS$$ $$(30)^2 = 2(10)S$$ $$900 = 20S$$ $$S = 45 \, \text{m}$$

So, the required distance is $45 \, \text{m}$.

The solution's first approach contains inconsistent intermediate steps, but the second approach correctly leads to the final answer $45$.