Originally Posted by

**galactus** I am including a diagram. I hope I am interpreting correctly.

We can use the horizontal and vertical component formulas.

horizontal: $\displaystyle x=(20cos(30))t$

vertical: $\displaystyle y=(20sin(30))t-\frac{1}{2}(9.8)t^{2}$

When the rock is level with the point it was thrown, then it's the point where y equals 0, we can find x at that time.

$\displaystyle 0=20sin(30)t-\frac{1}{2}(9.8)t^{2}$

Solving for t we find t=2.04 seconds to reach that point.

That means it's x coordinate is $\displaystyle 20cos(30)(2.04)=35.35 \;\ m$

The stone is at (35.35,0) at 2.04 seconds.

Now, we have a triangle with angle 45 and side 35.35

So, the y component is the same, y=-35.35 feet

Plug this into the y component and solve for t.

$\displaystyle -35.35=20sin(30)t-\frac{1}{2}(9.8)t^{2}$

And we find **t=3.89 seconds**

Perhaps TopsQuark will check me out.