Kinematics problem involving two objects

Question

Two objects are thrown up. Both are thrown up at 16 m/s, but one of the them is thrown up 2 seconds later after the first object is thrown. When and where do the objects meet?

 

I'm ending up with an unsolvable equation

Answer 1

Hi Gnarls,

 

I'm assuming that these objects are being thrown upwards from the surface of the earth, so they're affected by gravity?

 

Here's what I've got. Let's compare.

 

For the first object, I have the kinematic equation:

 

     y₁ = (16m/s)t - (4.9m/s²)t²

 

For the second object, I have the same kinematic equation, except it's delayed by two seconds, so I'll replace t with (t - 2s):

 

     y₂ = (16m/s)(t - 2s) - (4.9m/s²)(t - 2s)²

 

We want to know when and where the two objects meet, so let's set y₁ = y₂ and solve for t:

 

     (16m/s)t - (4.9m/s²)t² = (16m/s)(t - 2s) - (4.9m/s²)(t - 2s)²

     (16m/s)t - (4.9m/s²)t² = (16m/s)t - (32m) - (4.9m/s²)t² + (19.6m/s)t - (19.6m)

 

The first and second terms on the left hand side will cancel out the first and third terms on the right hand side:

 

     0 = -(32m) + (19.6m/s)t - (19.6m)

     0 = (19.6m/s)t - 51.6m

 

Solve for t:

 

     t = 51.6m / (19.6m/s)

     t = 2.632653s

 

So that's when they meet. Let's plug t = 2.632653s back into either of the two kinematic equations to find where they meet:

 

     y₁ = (16m/s)(2.632653s) - (4.9m/s²)(2.632653s)²

     y₁ = 8.161224m.

 

     y₂ = (16m/s)(2.632653s - 2s) - (4.9m/s²)(2.632653s - 2s)²

     y₂ = 8.161224m.

 

Since we've got the same answer from both kinematic equations, we can be pretty sure that this is the right answer.

 

Hope this has helped!