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?

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.

To answer these questions, we can use basic principles of physics and kinematics. Let's break it down step by step:

1. Initial velocity:
The initial velocity of an object can be calculated using the formula:
v = u + at
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken. In this case, the ball is thrown vertically, so the acceleration will be due to gravity (9.8 m/s^2). Since the ball is thrown vertically upwards, its final velocity will be zero at the maximum height. Therefore, we can rewrite the formula as:
0 = u - (9.8 m/s^2) * t
Solving for u:
u = (9.8 m/s^2) * t

2. Time to hit the ceiling:
To find the time it takes for the ball to hit the ceiling, we can use the equation for displacement of an object under constant acceleration:
s = ut + (1/2) * a * t^2
In this case, the initial displacement (s) is 2.4 m, the initial velocity (u) is unknown, acceleration (a) is -9.8 m/s^2 (negative because the ball is moving in the opposite direction to the acceleration due to gravity), and the final displacement (s) is 0 m. We can rewrite the formula as:
0 = u * t + (1/2) * (-9.8 m/s^2) * t^2
Solving this equation will give us the time it takes for the ball to hit the ceiling (t).

3. Time for the two balls to pass each other:
Since both balls are thrown with the same initial velocity, their motion can be treated independently. Therefore, the time it takes for the two balls to pass each other will be the same as the time it takes for the second ball to reach the height of the first ball at the time they cross paths. We can use the displacement equation again to calculate this time.

4. Height above the juggler's hand when the balls pass each other:
To find the height at which the balls pass each other, you can use the equation for displacement again. The displacement of the first ball can be calculated at the time they cross paths, and then the height above the juggler's hand can be found by subtracting the height of the juggler's hand from the calculated displacement.

By plugging in the known values and solving the relevant equations, you can find the answers to the specific values in this scenario.