A bouncy ball is dropped from a height of 144 feet. The function h=−16x^2+144 gives the height, h, of the ball after x seconds. When does the ball hit the ground?

3 Answers
Feb 24, 2018

The ball will hit the ground in 3 secs.

Explanation:

16x^2 must equal +144 so that h = 0.

In other words, when the ball hits the ground, the distance above the ground will be zero.

That occurs 3 seconds after the ball was released.

0=-16x^2 +144

16x^2 =144

Feb 24, 2018

3 seconds

Explanation:

h(x)=-16x^2+144 => the ball hits the ground at h = 0:
-16x^2+144=0
-16(x^2-9)=0
x^2=9
x=+-3 => reject the negative time, thus:
x=3 the ball hits the ground 3 seconds after it was dropped

Feb 24, 2018

The ball hits the ground after 3 seconds.

Explanation:

When the ball is dropped, the height will get less and less. When it hits the ground the height will be 0

0 = -16x^2 +144" "larr now solve the equation,

You can isolate the x term:

0 = -x^2 +9" "larr div 16

x^2 = 9

x =+-sqrt 9

color(blue)(x = +3) or x=-3

You could also factorise:

0 =144-16x^2" "larr difference of two squares.

0 = 9-x^2" "larr div 16

0 = (3+x)(3-x)

x= -3 or color(blue)(x=+3)

In each case reject -3 because the time cannot be negative,