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# math question

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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 ballâ€™s maximum height? Round to the nearest hundredth, if necessary.

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

#1
+6538
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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
Sort:

#1
+6538
+2

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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