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# help

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The equation y = -16t^2 - 60t + 54\$ describes the height (in feet) of a ball thrown downward at 60 feet per second from a height of 54 feet from the ground. In how many seconds will the ball hit the ground? Express your answer as a decimal rounded to the nearest hundredth.

May 16, 2019

#1
+4

height  =  -16t2  - 60t + 54     , where  t  is the number of seconds after the ball is released

In how many seconds will the ball hit the ground?

In other words, what is  t  when the height is zero?

Plug in  0  for the height and solve for  t .

0  =  -16t2  - 60t + 54

We can divide both sides of the equation by  -2

0  =  8t2 + 30t - 27

Let's split  30t  into two terms such that their coefficients multiply to  -216

0  =  8t2 - 6t + 36t - 27

Factor  2t  out of the first two terms and factor  9  out of the last two terms

0  =  2t(4t - 3) + 9(4t - 3)

Factor  (4t - 3)  out of both remaining terms

0  =  (4t - 3)(2t + 9)

Set each factor equal to  0  and solve for  t

 4t - 3  =  0 _______or_______ 2t + 9  =  0 4t  =  3 2t  =  -9 t  =  3/4 t  =  -9/2 t  =  0.75 t  =  -4.5

We want the number of seconds after the ball is released, so we only want the positive solution.

The ball will hit the ground  0.75  seconds.

Here's a graph: https://www.desmos.com/calculator/kznk4jayc4

May 16, 2019
#2
-5

Can you use the quadratic formula to solve this?

ProffesorNobody  May 16, 2019
#3
+3

Yep! You can use the quadratic formula to solve   -16t2  - 60t + 54  =  0   for  t , like this:

$$t \,=\, {-(-60) \,\pm \,\sqrt{(-60)^2-4(-16)(54)} \over 2(-16)}\,=\,{60\, \pm \,\sqrt{7056} \over -32}\,=\, {60 \,\pm\, 84 \over -32}\\~\\ \ \\ \begin{array}\ t\,=\,\frac{60+84}{-32}&\qquad\text{or}\qquad& t\,=\,\frac{60-84}{-32}\\~\\ t\,=\,\frac{144}{-32}&\qquad\text{or}\qquad& t\,=\,\frac{-24}{-32}\\~\\ t\,=\,-4.5&\qquad\text{or}\qquad& t\,=\,0.75 \end{array}$$ hectictar  May 16, 2019
edited by hectictar  May 16, 2019