h= (-16t^2 + 28t + 7)
find the derivative to get time
h' = (-32t + 28)
When the ball reaches its maximum height, its
vertical velocity (h’) will be 0
0 = -32t + 28
-32t=-28
-28/-32 = 0.875
seconds
The ball reaches its maximum height in 0.875
seconds after being thrown; plug and chug and:
h= (-16(0.875)^2 + 28(0.875) + 7) = 19.25 feet
Max height = 19.25 feet, time = 0.88 seconds
