Question

a juggler throws a ball into the air, releasing it 5 feet above ground with an initial velocity if 15 fps. she catches the ball in thr other hand 5 feet above the ground. How long was the ball in the air?

Answer 1

Problems of this sort are frequently found in physics.  If you study calculus or physics you'll learn how to create the equation representing the velocity of an object in flight.

Here, you don't need to calculate velocity, but rather time.  Start with this equation:

v = v0 + a t^2, where v is the velocity at time t, v0 is the initial velocity, a is the acceleration due to gravity (denoted by g instead of a), and t is the elapsed time.

You are told that v0 is 15 ft/sec.  Set v = to 0, as the ball stops moving for the tiniest instant at the top of its trajectory.  Use g = - 32 ft (per second squared).

Then  0 = 15 ft/sec - 32 [ft/(seconds squared)] t.

Solve this for t.  This is the time required for the ball to come to a complete stop at the top of its trajectory.  

Finally, multiply this time by 2, since the ball begins to fall and returns to its original height.