Question

How high will a baseball that is thrown up into the air at 48 feet per second go if it starts its flight at a height 10 feet above the ground?

Answer 1

height = -16t^2 + 48t + 10

d(height)/dt = -32t + 48
= 0 at max height
32t=48
t = 48/32 = 1.5

so at 1.5 seconds,
height = -16(1.5)^2 + 48(1.5) + 10 = 59.5 ft

check my arithmetic