Question

Ball A is thrown vertically upwards with a velocity of v0 . Ball B is thrown upwards from the same point with the same velocity t seconds later.

Part A

Determine the elapsed time t<2v0/g from the instant ball A is thrown to when the balls pass each other .

Express your answer in terms of some or all of the variables v0, t, and the acceleration due to gravity g .

Answer 1

Answer:

The answer is $\tau = \frac{v_o}{g} + \frac{t}{2}$

Explanation:

If the ball is thrown vertically, the equation of its position is

$y(\tau) = y_0 + v_0\tau - \frac{1}{2}g\tau^2$

So setting our coordinate system in the position of throwing $(y_0=0)$, the equation for A is

$y_A(\tau) = v_0\tau - \frac{1}{2}g\tau^2$

and for B

$y_B(\tau') = v_0\tau' - \frac{1}{2}g{\tau'}^2$

where

$\tau' = \tau - t$

due to the delay of the throwing between A and B, "t"

$t = \tau - \tau'$.

Now, the balls passing each other means that

$y_B(\tau') = y_A(\tau)$

then

$v_0(\tau - t) - \frac{1}{2}g(\tau - t)^2 = v_0\tau - \frac{1}{2}g\tau^2$

cancelling some terms...

$-v_0t + g\tau t - \frac{1}{2}gt^2 = 0$

so

$\tau = \frac{v_o}{g} + \frac{t}{2}$