Aball is thrown into the air with an initial upward velocity of 46 ft/s. it's height h in feet after t seconds is given by the function h=-16t squared plus 46t plus 6

Given the function 
h = -16t^2 + 46t + 6
and the initial velocity of 46 ft/s
to solve for the maximum height, first we must solve the time it reaches the maximum height. this done by getting the first derivative and equate it to zero

h = -16t^2 + 46t + 6
dh/dt = -32t + 46
-32t + 46 = 0
t = 1.4375 s it take to reach the maximum height

so the maximum height is
h = -16t^2 + 46t + 6
h = -16(1.4375^2) + 46(1.4375) + 6
h = 39.0625 ft

50.2450.24 in squared