In a movie production, a stunt person must leap from a balcony of one building to a balcony 3.0 m lower on another building. If the buildings are 2.0 m apart, what is the minimum horizontal velocity the stunt person must have to accomplish the jump? Assume no air resistance and that ay = −g = −9.81 m/s2 . (Ans. 2.6m/s) PLS SHOW WORK

You are looking for Vx0 (initial velocity in X direction) use this formula
remember there is no acceleration in x direction, only Y

X = x0 + Vx0(t) + (1/2)at^2
X= 0 + Vx0 (t) + 0
x= Vx0 (t)
Vx0 = x/t

what we are missing is Time, so u can get this from using the Y formula to find time
Remember there is no velocity in Y direction

Y=y0 + Vy0(t) - (1/2)gt^2
y= 0 +0 - (1/2)gt^2
(3.0 m *2) / 9.80m/s/s = t^2
0.61224 = t^2
t = 0.78 s

now plug this time back into this formula from above
Vx0 = x/t
= 2.0m / 0.78s
=2.6