Question

Dante makes a pass during a basketball game. The height y of the basketball after x seconds can be modeled by y= −16x2+34x+3 .

If no one catches the pass, how long will it take for the basketball to hit the ground? Round to the nearest tenth of a second.

Answer 1

Answer: 2.2 seconds

Step-by-step explanation:

Since the basketball hit the ground, y should equal 0.

So -16x^2+34x+3=0

Use the quadratic formula x = (-b ± square root(b^2 - 4ac)) / 2a

In this case, a = -16, b = 34, and c = 3

So x = (-34 ± sqrt(34^2 - 4(-16)(3))) / 2(-16)

After simplifying, we get x≈-0.08 or x≈2.2

We can discard the negative value. So the answer is 2.2

Answer 2

Explanation:

h(x) = y = -16x² + 34x + 3

I assume the height is measured and calculated in feet.

anyway, to hit the ground means the height is 0.

so, we need to find the value of x (number of seconds) for which the function result is 0.

h(x) = 0

0 = -16x² + 34x + 3

a quadratic equation

ax² + bx + c = 0

has the general solutions

x = (-b ± sqrt(b² - 4ac))/(2a)

in our case

a = -16

b = 34

c = 3

x = (-34 ± sqrt(34² - 4×-16×3))/(2×-16) =

= (-34 ± sqrt(1156 + 192))/-32 =

= (-34 ± sqrt(1348))/-32 =

= (-34 ± 2×sqrt(337))/-32 =

= (-17 ± sqrt(337))/-16

x1 = (-17 + sqrt(337))/-16 = -0.084847484... seconds

x2 = (-17 - sqrt(337))/-16 = 2.209847484... seconds

the negative solution x1 does not make any sense, so, x2 is our valid solution.

the ball will hit the ground after about 2.2 seconds.