A man standing on the edge of a cliff throws a stone straight up with initial speed u and then throws another stone straight down with the same initial speed and from the same position. Find the ratio of the speed the stones would have attained when they hit the ground at the base of the cliff :

Question

A man standing on the edge of a cliff throws a stone straight up with initial speed u and then throws another stone straight down with the same initial speed and from the same position. Find the ratio of the speed the stones would have attained when they hit the ground at the base of the cliff : (A) 4: 1 (B) 1: 3 (C) 1: 1 (D) 1 :2

Tardigrade Admin · Accepted Answer

As the stone thrown vertically up will come back to the point of projection with same speed, both the stones will move downward with same initial velocity, so both will hit the ground with velocity

v^{2} = u^{2} + 2 g h

i.e.,

v = (u^{2} + 2 g h)

So, the ratio of speeds attained when they hit the ground is

1 : 1

Q. A man standing on the edge of a cliff throws a stone straight up with initial speed u and then throws another stone straight down with the same initial speed and from the same position. Find the ratio of the speed the stones would have attained when they hit the ground at the base of the cliff :

$4 : 1$

$1 : 3$

$1 : 1$

$1 : 2$

Q. A man standing on the edge of a cliff throws a stone straight up with initial speed u and then throws another stone straight down with the same initial speed and from the same position. Find the ratio of the speed the stones would have attained when they hit the ground at the base of the cliff :

4:1

1:3

1:1

1:2

As the stone thrown vertically up will come back to the point of projection with same speed, both the stones will move downward with same initial velocity, so both will hit the ground with velocity v2=u2+2gh i.e., v=(u2+2gh)​ So, the ratio of speeds attained when they hit the ground is 1:1

$4 : 1$

$1 : 3$

$1 : 1$

$1 : 2$