Aball 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.

height of the ball = h(t) = -16t2 + 20t + 2 The maximum value for h(t) will occur when t = -20/2(-16) = 5/8 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.

No i refuse to tell you the answer but the estimate is 10