Question

A body falling under gravity covers two points A and B separated by 80 m in 2s. The distance of upper point A from the starting point is ________ m (use g = 10 ms–2)

Answer 1

Given:
- Distance between A and B = $80 \, \text{m}$,
- $t = 2 \, \text{s}$,
- $g = 10 \, \text{m/s}^2$.

Using the equation of motion:

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

For motion from A to B:

$-80 = v_1 t - \frac{1}{2} g t^2$

Substituting values:

$-80 = v_1 \cdot 2 - \frac{1}{2} \cdot 10 \cdot 2^2$

$-80 = 2v_1 - 20$

$-60 = 2v_1 \implies v_1 = -30 \, \text{m/s}.$

For motion from 0 to A:

Using the equation:

$v_1^2 = u^2 + 2gS$

$30^2 = 0 + 2 \cdot 10 \cdot S$

$900 = 20S \implies S = 45 \, \text{m}.$

The Correct answer is: 45 m