okay:
First of all, i recommend you try and draw the problem, as it always help to understand.
So, you need two equations, one for the horizontal movement, and one for the vertical, since the object will move both horizontally (it has a horizontal speed), and vertically (it's falling)
it's initial position can be given by the vector (x,y) = (0,5)
Horizontal :
x = x_{0} + v_{x} \times t + \frac{1}{2} \times a \times t^{2}
Vertical:
y = y_{0} + v_{y} \times t + \frac{1}{2} \times g \times t^{2}
g = 9.81
first you want to know at what time it hits the ground, so, you just need the vertical equation. It hits the ground when y = 0.
y_{0} , is the initial vertical position. Since it's 10m above the ground, it equals 10.
v_{y} = 0, because there is no initial velocity downwards or upwards, and the accelaration is, as stated, 9.81. Since the movement is downwards, the accelaration is negative (of course this depends on the referential you choose, but generally speaking, it's simpler this way)
Therefor :
Vertical:
0 = 10 + 0 \times t - \frac{1}{2} \times g \times t^{2}
t = \sqrt{\frac{10 \times 2}{g}}
solve that, and you have the time it hits the floor.
Then you need to know the distance it traveled.
Now, you need the horizontal equation.
We know it's initial horizontal position is 0, initial velocity 5, and there is no accelaration given, so:
x = 0 + 5 \times t + 0
Now, take the t you got from your first equation, and substitute it here. Now you do the math.
This is the general basis for projectile based problems: You should remember it.
You have to consider always an equation for each axis of movement, consider the velocity in each axis, the initial position in each axis, and the accelaration.