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A ball is thrown vertically upward from the ledge of a building 75 feet above ground. The ball's height $h$ in feet above the ground at time $t$ in seconds is given by the equation$$h=-16 t^{2}+80 t+75.$$(a) how long does it take the ball to reach a height of 90 feet? (b) How long before the ball is back to its original position (at 75 feet)? (c) How long before the ball hits the ground? Give each answer to the nearest one-thousandth of a second.

(a) 0.195 sec or 4.805 sec(b) $5 \sec (\mathrm{c}) 5.807 \mathrm{sec}$

Algebra

Chapter 0

Reviewing the Basics

Section 4

The Quadratic Formula and Applications

Equations and Inequalities

Oregon State University

Harvey Mudd College

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mhm. The height is a quadratic function of time in part A. It asks How now that it takes the board to reach a height of 90 ft. So we just decided height to 90 ft. And then we solve this quadratic equation. First, we subtract the right hand side from both sides of this Ukraine and the quadratic equation becomes 16 K squared miners 80 t Plus 15 equals zero. We can identify the coefficient as A Echoes 16 B, Eco's miners 80 And say he calls 15 and then we substitute A B and C with their values into a formula which gives us the denominator is two times 16. The numerator is Eddie plus um miners square root out minus 80 Squared miners four times 16 times 15 which echoes 80 plus some miners a square road of Fire saw than the 440 or 32 With vehicles 10 plus on minors. The Square Rhoda of 85 Werfel. And if we use a calculator, their approximate values .195 and four points 805 Notes that there are two solutions. They're satisfied this question because first the board will go upward and then it reaches the highest heart and then it will go down. So maybe 90 fit is here. So the ball will pass through the 90 ft Twice. And that is why there are two solutions for part B. It asks how now before the bore is right to its original position. Okay, Height of the original position is 75 ft. So we just beside a height to 75 ft. Still, we subtract the right hand side from both sides of the secret and it becomes 16 T squared, miners 80 equals zero. And then they can right this equating as 16 t Times T -5 Echo zero. Then we can see that there are two solutions TV Angle zero. Anti miracles five. Hence the answer in 50 is just the original time. That is at T equals zero. The boy is at its original position which is up views for piracy. Move on to know how long before the board hits the ground. The height above ground is there? Oh so we just aside the height to zero. Still we subtract the right hand side from both sides of the issue crater and it becomes 16 T squared. Class A. T. T. Sorry should be miners Minors 75 equals zero. Then they can identify the confessions as a equals 16 Vehicles, Miners 80 and Singles Minors 75. Then we substitute A. B. And C. V. Their values into this form. The dinner major is two times 16. The numerator is 80 plus some miners. The square root miners 80 squared -4 times 16 times -75, which it calls 80 plus on minors Square root health. You haven't saw them to 200 over 32 which echoes 10 plus miners five times the square root out seven awful. If we use a calculator, they can obtain the Approximate value of these two solutions. The first is erosion is miners points 807 The second, so Russia has 5.807. Notes out. The time cannot be negative and hence, which has a eliminates the first solution. And the second solution is that result we want

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