Question

If a juggler throws a ball (10cm in diameter) 2.4m vertically above the level of his hand to the ceiling;

1. What is it's initial velocity? I work out 0m/s?

2. How long does it take for the ball to hit the ceiling?

3. If he throws a second ball at the same initial velocity as the first, how long after the second ball is thrown do the two balls pass each other?

4. When they do pass each other how far above the jugglers hands are they?

Answer 1

1. Vf^2 = Vo^2 + 2(-9.8)2.4 = 0,
Vo^2 - 47.04 = 0,
Vo^2 = 47.04,
Vo = 6.9m/s.

2. Vf = Vo + gt = 0,
6.9 - 9.8t = 0,
-9.8t = -6.9,
t = 0.70s.