# Calculate the change in time between two throws

1. Nov 9, 2008

### Altairs

1. The problem statement, all variables and given/known data
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2. Relevant equations

In picture attached

3. The attempt at a solution

The confusion is that why will the final time of the two balls be the same i.e. 't'...The time taken by the two balls can never be the same....Am confused at the same final time 't'..Please explain.

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2. Nov 9, 2008

### tiny-tim

Hi Altairs!
Because the boy throws the first ball at t = 0, and the second ball at t = ∆t.

That's why (t - ∆t) keeps appearing in the equations!

3. Nov 9, 2008

### Altairs

My pooint is why the same 't' ?

4. Nov 9, 2008

### tiny-tim

It's as if the boy has two stop-watches, one for each ball, each starting at zero when that ball is thrown.

The first watch measures time t, starting at t = 0, and so it shows t = ∆t when the second ball is thrown.

The second watch measures time u, starting at u = 0, which is the same as t = ∆t.

From then on, the second watch is always ∆t slower than the first watch … in other words, u = t - ∆t.

The two formulas do not use the same t … they use t and u …

but you can convert u into t.