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

 Jan 25, 2021

Best Answer 

 #1
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When the ball hits the ground, h will be zero

sooooo    0 = -16t^2 + 60t + 54      Use quadratic formula to solve for t  (throuw out negative value)  

                       a = -16      b=60     c = 54

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

 

\(t = {-60 \pm \sqrt{60^2-4(-16)(54)} \over 2(-16)}\)

 Jan 25, 2021

1+0 Answers

 #1
avatar
+1
Best Answer

When the ball hits the ground, h will be zero

sooooo    0 = -16t^2 + 60t + 54      Use quadratic formula to solve for t  (throuw out negative value)  

                       a = -16      b=60     c = 54

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

 

\(t = {-60 \pm \sqrt{60^2-4(-16)(54)} \over 2(-16)}\)

Guest Jan 25, 2021

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