A ball is thrown from a height of

123

feet with an initial downward velocity of

11/fts

. The ball's height

h

(in feet) after

t

seconds is given by the following.

=h-123-11t16t2

How long after the ball is thrown does it hit the ground?

Define

g = 9.8 32.2 ft/s², the acceleration due to gravity.

Refer to the diagram shown below.

The initial height at 123 feet above ground is the reference position. Therefore the ground is at a height of - 123 ft, measured upward.

Because the initial upward velocity is - 11 ft/s, the height at time t seconds is

h (t) = - 11t - (1/2) gt²

or

h (t) = - 11t - 16.1t²

When the ball hits the ground, h = - 123.

Therefore

-11t - 16.1t² = - 123

11t + 16.1t² = 123

16.1t² + 11t - 123 = 0

t² + 0.6832t - 7.64 = 0

Solve with the quadratic formula.

t = (1/2) [-0.6832 + / - √ (0.4668 + 30.56) ] = 2.4435 or - 3.1267 s

Reject the negative answer.

The ball strikes the ground after 2.44 seconds.

Answer: 2.44 s