A ball is thrown vertically upward from the top of a buildin

A ball is thrown vertically upward from the top of a building 128 feet tall with an initial velocity of 112 feet persecond. the distance s (in feet) of the ball from the ground after t seconds is s =128 +112t-16t². after how many seconds will the ball pass the top of the buildings.

Solution

s = 128+112t-16t^2
128 = 128+112t-16t^2
0 = 112t-16t^2
0 = t(112-16t)

t = 0
or
16t = 112
t = 112/16
t = 7

Answer:
Obviously the ball is at building height at 0 seconds,
but the answer to the question is that it will pass
that same height after 7 seconds.