Find the time(s) when the ball's height is 48 feet.

Question

Dana is going to throw a ball from the top floor of her middle school. The function h(t)= -16t²+32t+48 models the height, h, of the ball above the ground as a function of time t.

Find the time(s) when the ball's height is 48 feet.

The ball has a height of 48 feet when t = ____ seconds. (If there is more than one answer, express your answers separated by commas)

The ball hits the ground at t = ____ seconds.

Answer 1

h(t)= -16t² + 32t + 48

Height = 48 feet

To find when ball has a height of 48 feet,

set h(t) = 48.

-16t² + 32t + 48 = 48 means that

-16t² + 32t = 0

So -16t(t – 2) = 0

The two solutions are t = 0 and t = 2

Height = 0 feet

To find when ball hits the ground,

set h(t) = 0.

-16t² + 32t + 48 = 0 means that

-16(t² – 2t – 3) = 16(t – 3)(t + 2)

So solutions come from

t – 3 = 0 which means t = 3

t + 2 = 0 which would mean t = –2

Since it's not possible for t to be less than zero (ar least not without time travel), we discard the second solution.

Answers

The ball has a height of 48 feet first at t = 0 seconds, then again at t = 2 seconds.

The ball hits the ground at t = 3 seconds.

Answer 2

h(t) = -16t² + 32t + 48 = 48

-16t² + 32t = 0 -16t(t - 2) = 0 So, t = 0 or t = 2.

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The ball hits the ground when h(t) = 0

-16t² + 32t + 48 = 0

-16(t² - 2t - 3) = 0

-16(t - 3)(t + 1) = 0

t = 3 or t = -1

Since time can't be negative, t = 3.

The ball hits the ground when t = 3 sec.