Sarah G.

asked • 12/15/17# A ball is thrown downward from the top of a 130-foot building with an initial velocity of 22 feet per second. The height of the ball h after t seconds is given

A ball is thrown downward from the top of a 130-foot building with an initial velocity of 22 feet per second. The height of the ball h after t seconds is given by the equation .h= −16t2−22t+130. How long after the ball is thrown will it strike the ground?

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## 2 Answers By Expert Tutors

Mark O. answered • 12/15/17

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Hi Sarah,

Let y be the height above the ground for the ball. We can then write the equation as

y = -16t

^{2}-22t + 130If t = 0, then y = 130 feet as we are told.

We want to know when the ball strikes the ground, or when y = 0. So, set y = 0 and solve the quadratic for t.

-16t

^{2}-22t + 130 = 0We can divide the equation through by -2 to get

8t

^{2}+ 11t - 65 = 0Unfortunately, there is no easy way to factor this equation. We go to the quadratic formula.

t = [-11 +/- [11

^{2}- 4(8)(-65)]^{1/2}}/{(2)(8)}We can simplify this as

t = [-11 +/- √2201]/16

√2201 > 11, so we only keep the positive root, since time must be positive.

So,

t = [-11 + √2201]/16 = 2.244 sec

Andrew M. answered • 12/15/17

Tutor

New to Wyzant
Mathematics - Algebra a Specialty / F.I.T. Grad - B.S. w/Honors

h(t) = -16t

^{2}- 22t + 130This equation will find the height at time t.

We want the value of t when h(t) = 0

-16t

^{2}- 22t + 130 = 0We can factor out a 2 since all are even numbers...

Better yet, just to make the coefficient of t

^{2}positive,let's factor out a -2

-2(8t

^{2}+ 11t - 65) = 08t

^{2}+ 11t - 65 = 0Now we have a quadratic equation of at

^{2}+ bt + c = 0t = [-b ±√(b

^{2}-4ac)]/2a with a=8, b=11, c=-65t = [-11 ±√(11

^{2}-4(8)(-65))]/[2(8)]t = [-11 ±√(121 + 2080)]/16

t = [-11 ±√2201]/16

t = (-11 ± 46.915)/16

Since our time will not be negative, we can discount the (-11-46.915)/16

t = (-11 + 46.915)/16 = 35.915/16 ≈ 2.2447 seconds

Round to as many decimals as needed

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Andrew M.

12/15/17