Question

A juggler throws a ball into the air, releasing it 5 feet above the ground with an initial velocity of

15 ft/sec. She catches the ball with her other hand when the ball returns to 5 feet above the ground. If the equation h=-16t^2 +
15t gives the path of the ball from hand to hand, find how long the ball is in the air

Answer 1

The ball is in the air for 0.9375 seconds. Please check the explanation below to help you remember the steps.

Explanation:

The equation h = -16t² + 15t + 5 represents the height of the ball at any time t. We want to find when the ball returns to a height of 5 feet, so we set h = 5 and solve for t.

5 = -16t² + 15t + 5

Subtract 5 from both sides to set the equation to 0:

0 - -16t² + 15t

Factor out t:

0 = t(-16t +15)

Setting each factor equal to zero gives the solutions:

t = 0 and -16t + 15 = 0

Solving the second equation for t gives:

t = 15/16

So, the ball is in the air for 0.9375 seconds.