Before you begin, you need to first figure out what your problem gives you and what you need to look for.
Given:
vix=10m/s
dy= 20m
What you are looking for is horizontal distance or dx.
The formula for vertical distance is:
[tex]dx=vi_{x}t[/tex]
Where:
dx= horizontal distance
vix= initial horizontal velocity
t=time
But when you look at your problem, you do not have time. Now for time, you need to use the formula:
[tex]t= \sqrt{ \frac{2dy}{g} } [/tex]
g in this equation is a constant. It is the acceleration due to gravity which has the value of 9.8m/s^2.
Now we first need to get time:
[tex]t= \sqrt{ \frac{2dy}{g} } [/tex]
[tex]t= \sqrt{ \frac{2(20m}{9.8m/s^{2}} } [/tex]
[tex]t= \sqrt{ 4.08163265} [/tex]
[tex]t=2.02s [/tex]
Now that you have time, all you need to do is insert that into our equation for horizontal distance.Â
[tex]dx=vi_{x}t[/tex]
[tex]dx=(10m/s)(2.02s)[/tex]
[tex]dx=20.20m[/tex]
The horizontal distance of the rock will be 20.20m.
