Answer:
We know that the second equation of motion is
S= ut + 1/2a²
And S is displacement and u is initial velocity
So in the case of Haley lets take downwards as positive Y-axis
S = 2h and
initial velocity = v
a = g (acceleration due to gravity = 9.8)
Substituting
2h = vt + 1/2gt²
And for Joe we take ownwards as positive Y-axis
S = h and
initial velocity = 0 (since the ball is dropped from rest)
a = g
h = 0x t + 1/2gt2²
t2= √ 2h/g
Now since both balls reach ground at same time: t1=t2
So
putting value of t2 in Hayley's equation:
2h= v(√2h/g) + 1/2 g( √2h/g)²
So v= √gh/2