Question

A ball is thrown into the air with an upward velocity of 28 ft/s. its height,h, in feet after t seconds is given by the function h=-16t^2+28t+7. what is the balls maximmum height? how long does it take the ball to reach its maximum height? round to the nearest hundredth, if necessary.

PLZZ HELP ME.........ASAP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Answer 1

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