Respuesta :

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.