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A ball is thrown into the air with an upward velocity of 20 feet per second. Its height, h, in feet after t seconds is given by the function h of t equals negative 16t squared plus 20t plus 2 . How long does it take the ball to reach its maximum height? What is the ball’s maximum height? Round to the nearest hundredth, if necessary.

Guest Nov 15, 2017
edited by Guest  Nov 15, 2017

Best Answer 

 #1
avatar+5589 
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height of the ball   =   h(t)   =   -16t2 + 20t + 2

 

The maximum value for  h(t)  will occur when  t  =  \(\frac{-20}{2(-16)}\)   =   \(\frac58\)

 

It will take  5/8 of a second,  or  0.625 seconds, for the ball to reach its maximum height.

 

To find the ball's maximum height, plug in  5/8  for  t .

 

h( 5/8 )  =  -16( 5/8 )2  +  20( 5/8 ) + 2

 

h( 5/8 )  =  -16( 25/64 )  +  50/4  +  2

 

h( 5/8 )  =  -25/4  +  50/4  + 8/4

 

h( 5/8 )  =  33/4

 

The maximum height is  33/4  feet,  or  8.25  feet.

 

Here's a graph.

hectictar  Nov 15, 2017
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1+0 Answers

 #1
avatar+5589 
+2
Best Answer

height of the ball   =   h(t)   =   -16t2 + 20t + 2

 

The maximum value for  h(t)  will occur when  t  =  \(\frac{-20}{2(-16)}\)   =   \(\frac58\)

 

It will take  5/8 of a second,  or  0.625 seconds, for the ball to reach its maximum height.

 

To find the ball's maximum height, plug in  5/8  for  t .

 

h( 5/8 )  =  -16( 5/8 )2  +  20( 5/8 ) + 2

 

h( 5/8 )  =  -16( 25/64 )  +  50/4  +  2

 

h( 5/8 )  =  -25/4  +  50/4  + 8/4

 

h( 5/8 )  =  33/4

 

The maximum height is  33/4  feet,  or  8.25  feet.

 

Here's a graph.

hectictar  Nov 15, 2017

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