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**+2 **

height of the ball = h(t) = -16t^{2} + 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.

hectictar
Nov 15, 2017

#1**+2 **

Best Answer

height of the ball = h(t) = -16t^{2} + 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.

hectictar
Nov 15, 2017