A ball is thrown horizontally from the roof of a building 50 m tall and lands 35 m from the base. What was the

A ball is thrown horizontally from the roof of a building 50 m tall and lands 35 m from the base. What was the ball's initial speed

vectors are handled separately

the up down vector of gravity is 90 degs or perpendicular to the horizontal motion vector
so we can take each separate

since the throw was horizontal it had 0 vertical speed
we know the height 50 m
y = 0.5 *a*t^2 is the distance formula for constant acceleration

50 = 0.5*9.8*t^2
solving for t
100/9.8 = t^2
t = 10/sqrt(9.8)

that is the time it takes to drop 50 m

therefore since it is travelling 35 m from the base at the same time

horizontal motion is constant velocity problem

x = v*t

since we know there is no change in speed as we assume 0 air resistance usually in these problems

solve for v

v = x/t
v = 35/(10/sqrt(9.8))
v = 35*sqrt(9.8)/10
v = 10.96 m/s
v = 11 m/s horizontally

this is all horizontal no vertical speed