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A classic physics problem states that if a projectile is shot vertically into the air with an initial velocity of 110 feet per second from an initial height of 59 feet off the ground, then the height of the projectile, [tex]h[/tex], in feet, [tex]t[/tex] seconds after it's shot, is given by the equation:

[tex]h(t) = -16t^2 + 110t + 59[/tex]

Find the two points in time when the object is 97 feet above the ground. Round your answers to the nearest hundredth of a second (two decimal places).

The object is 97 feet off the ground at the following times: [tex]t = [/tex] ____, ____.

(Enter your two answers separated by a comma.)

Answer :

The object is 97 feet off the ground at 0.06 seconds.

1: Set up the equation:

We are given the equation: h = 16t² + 110t + 59

We want to find the values of t when h = 97. Substitute this value into the equation:

97 = 16t² + 110t + 59

2: Solve the quadratic equation:

This is a quadratic equation in t. We can solve it by factoring, using the quadratic formula, or graphing. Here, we'll use the quadratic formula:

t = (-b ± √(b² - 4ac)) / 2a

where:

a = 16

b = 110

c = 59 - 97 = -38

Substitute these values into the formula:

t = (-110 ± √(110² - 4 * 16 * -38)) / 2 * 16

Calculate the discriminant (part under the square root):

110² - 4 * 16 * -38 = 12544

So, the square root is:

√12544 = 112

Therefore, the possible solutions for t are:

t1 = (-110 + 112) / 32 = 0.0625

t2 = (-110 - 112) / 32 = -6.9375

3: Round and answer:

Since the time cannot be negative, we reject the second solution (t2). Round the remaining solution to the nearest hundredth of a second:

t1 ≈ 0.06

Therefore, the object is 97 feet off the ground at 0.06 seconds after it is shot.

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