Question

two stones are dropped from the top of a tower at an interval of 3 seconds. the relative velocity of the first stone with respect to the second stone at an instant during their fall is​

Answer 1

Given:

Two stones are dropped from the top of a tower at an interval of 3 seconds.

To find:

The relative velocity of the first stone with respect to the second stone at an instant during their fall is​

Solution:

From given, we have,

two stones are dropped from the top of a tower at an interval of 3 seconds.

we use the formula,

v = u + at

v = 0 + gt

∴ v = gt

where g is the gravitational constant.

First stone,

V₁ = gt

Second stone,

V₂ = g(t - 3)

The relative velocity of the first stone with respect to the second stone at an instant during their fall is​

Relative velocity,

V = V₁ - V₂

= gt - g(t - 3)

= gt - gt + 3g

= 3g

Therefore, the relative velocity is 3g m/s

If considered, the gravitational constant, then we get,

Therefore, the relative velocity is 3 × 10 = 30 m/s

Answer 2

Explanation:

according to the question answer is given by me