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

In summary, to find the velocity of two stones when they meet, we use the equation V2 = u2 + 2as and set V1 = V2. This gives us a final velocity of u + ag. The time taken for the stones to meet is u/g.
  • #1
kalupahana
36
0

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.
:rolleyes:
 
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  • #2


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.
 

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

1. How do you calculate the velocity and time for two stones projected vertically with the same initial velocity?

The velocity and time for two stones projected vertically with the same initial velocity can be calculated using the formula: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity (9.8 m/s^2), and t is the time. The initial velocity for both stones will be the same, so the only difference will be the time it takes for each stone to reach the ground.

2. What is the initial velocity for the two stones in this project?

The initial velocity for the two stones in this project is the velocity at which the stones are projected upward. This can be measured using a speedometer or by calculating the velocity using the distance and time of projection.

3. How does the mass of the stones affect the velocity and time calculations?

The mass of the stones does not affect the velocity and time calculations in this scenario because both stones are projected with the same initial velocity. However, if one of the stones is significantly heavier than the other, it may experience more air resistance and reach the ground slightly later than the lighter stone.

4. Can the initial velocity be changed during the stones' projection?

No, the initial velocity cannot be changed during the stones' projection unless an external force is applied. In this scenario, we are assuming that both stones are projected with the same initial velocity and there are no external forces acting on them.

5. How can this experiment be modified to study the effects of air resistance on the stones' velocity and time?

To study the effects of air resistance on the stones' velocity and time, the experiment can be modified by projecting the stones at different angles and recording the time taken for each stone to reach the ground. This will allow for a comparison of the effects of air resistance on the stones' motion at different angles of projection. Additionally, the experiment can be conducted in different environments with varying air densities to gather more data on the impact of air resistance on the stones' velocity and time.

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