Solving for Velocity & Time for 2 Stones Projected Vertically with Same Initial Velocity

[kalupahana]

Homework Statement

A stone is projected vertically with a initial velocity u. Another stone projected from the same position verticaly with same velocity u after time t. Find the velocity of both stones when they meet together & time taken to meet.

Homework Equations

S = ut + 1/2 at²
v = u + at
V² = u² + 2as

The Attempt at a Solution

Max height = u/2g
time taken reach max height = u/g
Time taken to reach first stone ground = 2u/g

So I took time taken to meet two as

t + t/2 = 3t/2

Instance velocity of bath should be equal when they meet.
So
Using v = u + at
V= u + 3gt/2

But answer is given as gt/2, I cannot reach to this

& time is given as t/2 + u/g

I can take time if the t is time taken to reach the max point.
t= u/g
Then
Time = t/2 + u/g

Please help me to reach to this answer.

 

[anathema]

Hello!

To find the velocity of both stones when they meet, we need to use the equation V2 = u2 + 2as, where V is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement.

Since both stones are projected vertically from the same position, their displacement (s) will be the same. We can also assume that their acceleration is the same (due to gravity). So, we can set up the following equation:

V1^2 = u^2 + 2as

V2^2 = u^2 + 2as

Since we know that both stones will meet at a certain point, we can set V1 = V2 and solve for t (time taken to meet):

V1^2 = V2^2

u^2 + 2as = u^2 + 2as

2as = 2as

Since we know that s = u/2g (maximum height reached by the stone), we can substitute it in the equation above:

2a(u/2g) = 2a(u/2g)

a(u/g) = a(u/g)

Now, we can solve for t:

t = u/g

So, the time taken for both stones to meet is u/g.

To find the velocity at which they meet, we can substitute t = u/g in the equations we set up earlier:

V1^2 = u^2 + 2as

V2^2 = u^2 + 2as

V1^2 = u^2 + 2a(u/g)

V2^2 = u^2 + 2a(u/g)

Since V1 = V2, we can set the equations equal to each other:

V1^2 = V2^2

u^2 + 2a(u/g) = u^2 + 2a(u/g)

Solving for V1 (or V2) gives us:

V1 = V2 = u + ag

So, when the two stones meet, their velocity will be u + ag.

I hope this helps! Let me know if you have any further questions.