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Bezzdna [24]
2 years ago
7

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 .
Physics
1 answer:
Illusion [34]2 years ago
7 0

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}

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The angular acceleration when it stopping:

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The angular distance it covers when starting from rest:

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\theta_a = \frac{\omega^2}{2\alpha_a} = \frac{12.57^2}{2*1.257} = 62.8 rad

The angular distance it covers when coming to complete stop:

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\theta_o = \frac{-\omega^2}{2\alpha_o} = \frac{-12.57^2}{2*(-1.05)} = 75.4 rad

So the total angular distance it covers within 22 s is 62.8 + 75.4 = 138.23 rad or 138.23 / (2π) = 22 revolutions

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