A ball is thrown from an initial height of 6 feet with an initial upward velocity of 17 ft/s. The ball's height h (in feet) after t seconds is given by the following.


h=6+17t-16t^2


Find all values of t for which the ball's height is 10 feet.


Round your answer(s) to the nearest hundredth.
(If there is more than one answer, use the "or" button.)
t=___seconds

Answers

Answer 1

Answer:

6 + 29t - 16t^2 = 18

16t^2 - 29t + 12 = 0

t = (29 +- √(29^2 - 4*12*16))/32

t = (29 +- √73)/32

t  = 0.639249882958827, 1.1732501170411

Step-by-step explanation:

Answer 2

The values of t for which the ball's height is 10 feet are approximately 0.71 seconds or 0.35 seconds

The given equation for the height of the ball is a quadratic equation, which is solved using the quadratic formula to find the values of t.

The quadrature formula for ax² + bx + c = 0 is:

[tex]x = \frac{-b \pm \sqrt{(b^2- 4ac)}}{2a}[/tex]

Given that,

The initial height of the ball: 6 feet

The initial upward velocity of the ball: 17 ft/s

The equation for the ball's height after t seconds: h = 6 + 17t - 16t²

The goal is to find the values of t for which the ball's height is 10 feet.

To find the values of t when the ball's height is 10 feet,

Set the equation h = 10 and solve for t.

Plugging in the given equation h = 6 + 17t - 16t²

And substituting h with 10, we get:

10 = 6 + 17t - 16t²

Rearrange the equation to solve for t.

16t² - 17t + 4 = 0

To find the values of t, we can use the quadratic formula:

In this case,

a = 16,

b = -17, and

c = 4.

Plugging these values into the quadratic formula, we get:

[tex]t = \frac{17 \pm \sqrt{((-17)^2- 4 \times 16 \times 4)}}{2 \times 16}[/tex]

Simplifying this equation, we have:

[tex]t = \frac{(17 \pm \sqrt{(289 - 256)}}{32}[/tex]

[tex]t = \frac{(17 \pm \sqrt{33})}{32}[/tex]

t ≈ 0.710 or 0.351

Hence,

Rounding to the nearest hundredth, we get two possible values for t:

t ≈ 0.71 seconds or 0.35 seconds

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