Kinematics physics

Question

An object is projected vertically upwards with velocity u m.s^-1 , T seconds later another object is projected vertically upwards from the same point and with the same speed. Find, in terms of u , T and g , the further time which elapses before the objects collide

Answer 1

Hi Nona!

 

The key definition here is that two objects will collide if they are at the same position at the same time.  If they both start at the same position, which we presume, this means that they must have the same displacement at the same time.

 

For the first object, we know two vertical kinematic quantities:

 

a = -g (presuming down to be negative)

v_o = u

 

For the second object, we know:

 

a = -g

v_o = u

 

and the second object is launched T seconds after the first.

 

Since our collision condition involves displacement and time, it would be nice to use the kinematic equation involving those two quantities, plus the two we know.  That would be:

 

d = v_ot+(1/2)at²

 

For the first object, we can then write:

 

d₁ = ut₁+(1/2)(-g)(t₁) = ut₁-(1/2)gt₁²

 

For the second object:

 

d₂ = ut₂+(1/2)(-g)(t₂) = ut₂-(1/2)gt₂²

 

Now, as written, these are two equations with four unknowns:  d₁, d₂, t₁, and t₂.  However, we do have two conditions and relationships that will reduce this number of unknowns from 4 to 2:

 

Because we are looking for collision of objects launched from the same position, we need (as mentioned above):  d₁=d₂ (and thus, we can call both of them just "d")

 

Also, since the second object is launched T seconds after the first, we can write:

 

t₂ = t₁-T  (since, wherever the second object is on its path, it is where object 1 was T seconds before)

 

Thus, we can call t₁ "t" and then t₂ will be t-T.

 

The two equations above then become:

 

d = ut-(1/2)gt²

d = u(t-T)-(1/2)g(t-T)²

 

Now, this constitutes two equations with only 2 unknowns; and, if your number of unknowns equals your number of equations, you can solve for the unknowns.  In this case, you want to solve for t.

 

I would suggest setting the two right sides of the equations equal, and solving for the value of t that makes that equation true.  This will be the amount of time after object 1 is tossed up that the two objects will collide.

 

I hope this gets you started!  Just let me know if you want to check any work or talk about this more.