A juggler throws a ball straight up into the air. The ball remains in the air for a time before it lands back in the jugglers hand.

With what speed did the juggler throw the ball into the air?

Been a while since I took physics, but...
The total energy of the baal should remain equal. When the ball is a the peak of its height, it has 100% potential energy. When it is just about to leave the juggler's hand, it has 100% kinetic energy. These two values should be the same.

First, calculate the potential energy at the peak height. U=mgh where U=potential energy, m=mass of the ball, g=standard gravity (9.8m/s-squared), and h=height of the ball. This value should also equal the kinetic energy when the ball is leaving the juggler's hand.

For kinetic energy, E=0.5mv-squared where E=kinetic energy (equal to potential energy calculated above), m=mass of the ball, and v=velocity.

Mathematically
U=E
mgh=0.5mv-squared

since m is on both sides it can be eliminated (meaning you don't need to know the mass of the ball)

gh=0.5v-squared

solving for v (velocity)

v=square root of 2gh

Hope this makes sense.