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.
+9481 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.
+9481 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.
