Question

A ball is thrown from the top of a 50-ft building
with an upward velocity of 24 ft/s. When will it
reach its maximum height? How far above the
ground will it be?

Answer 1

Using your data,
h = -16t^2 + 24t + 50

This is a downwards opening parabola, so find its vertex and all the
mysteries will be revealed.

Answer 2

the model can be written this way:
height= vi*t + hi -16t^2 where hi=50ft; vi=24ft/sec
so this is a parabola. The zeroes can be found
0=23*t+50-16t^2
16t^2-23t-50=0

t=(23+-sqrt(23^2+4*16*50))/32

solving that, t=2.63, -1.15 so the max height occurs halfway between these zeroes
tmax=3.78/2-1.15= .74 sec
hmax=h(.74)=height= vi*t + hi -16t^2