A stone is dropped freely, while another is thrown vertically downward with an initial velocity at 2 m/s from the same point, simultaneously. The time required by them to have a distance of separation 22 m between them is:

Question

A stone is dropped freely, while another is thrown vertically downward with an initial velocity at 2 m/s from the same point, simultaneously. The time required by them to have a distance of separation 22 m between them is: (A) 11 s (B) 5.5 s (C) 44 s (D) 22 s

Tardigrade Admin · Accepted Answer

Given: distance of separation = 22 m u = 2 m/s From the formula, distance = velocity × time or s1-s2=ut 22=ut=2× t ∴ t=(22/2)=11 s

Q. A stone is dropped freely, while another is thrown vertically downward with an initial velocity at 2 m/s from the same point, simultaneously. The time required by them to have a distance of separation 22 m between them is:

11 s

5.5 s

44 s

22 s

Given: distance of separation = 22 m u = 2 m/s From the formula, distance = velocity $\times$ time or $s_{1} - s_{2} = u t$ $22 = u t = 2 \times t$ $∴$ $t = \frac{22}{2} = 11 s$

Q. A stone is dropped freely, while another is thrown vertically downward with an initial velocity at 2 m/s from the same point, simultaneously. The time required by them to have a distance of separation 22 m between them is:

11 s

5.5 s

44 s

22 s

Given: distance of separation = 22 m u = 2 m/s From the formula, distance = velocity × time or s1​−s2​=ut 22=ut=2×t ∴ t=222​=11s

Given: distance of separation = 22 m u = 2 m/s From the formula, distance = velocity $\times$ time or $s_{1} - s_{2} = u t$ $22 = u t = 2 \times t$ $∴$ $t = \frac{22}{2} = 11 s$