Answer:
At the same time.
Explanation:
In the first case ,
intial velocity = 0
displacement = -h
acceleration = -g
Using second equation of motion,
s = ut + .5a[tex]t^{2} \\[/tex]
- h = - 0.5g[tex]t^{2} \\[/tex]
t = [tex]\sqrt{\frac{2h}{g} }[/tex]
In the second case, consider only motion along y axis,
intial velocity = 0 ( all the velocity is along x axis)
displacement = -h ( height is same in both cases)
acceleration = -g
Using second equation of motion,
s = ut + .5a[tex]t^{2} \\[/tex]
- h = - 0.5g[tex]t^{2} \\[/tex]
t = [tex]\sqrt{\frac{2h}{g} }[/tex]
In both cases time is same.
Hence, they will reach the ground simultaneously.