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 .

Question

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carlostorres808 carlostorres808 · Answer 1 · 2019-09-10T06:28:16+02:00

Answer:

The answer is [tex]\tau = \frac{v_o}{g} + \frac{t}{2}[/tex]

Explanation:

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

[tex]y(\tau) = y_0 + v_0\tau - \frac{1}{2}g\tau^2[/tex]

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

[tex]y_A(\tau) = v_0\tau - \frac{1}{2}g\tau^2[/tex]

and for B

[tex]y_B(\tau') = v_0\tau' - \frac{1}{2}g{\tau'}^2[/tex]

where

[tex]\tau' = \tau - t[/tex]

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

[tex]t = \tau - \tau'[/tex].

Now, the balls passing each other means that

[tex]y_B(\tau') = y_A(\tau)[/tex]

then

[tex]v_0(\tau - t) - \frac{1}{2}g(\tau - t)^2 = v_0\tau - \frac{1}{2}g\tau^2 [/tex]

cancelling some terms...

[tex]-v_0t + g\tau t - \frac{1}{2}gt^2 = 0[/tex]

so

[tex]\tau = \frac{v_o}{g} + \frac{t}{2}[/tex]