Projectile Physics Problem: Understanding Rock's Horizontal Flight Time

[physucsc11]

Got stuck on this problem, would appreciate some help.

Problem: A rock is kicked horizontally at 15 m/s from a hill with a 45 deg slope.
How long does it take for the rock to hit the ground?

The book gives the answer in the back as 5.1 sec but I don't understand how
it gets this answer.

Thanks for any help.

 

[ozymandias]

Here is a hint:
Let's name the horizontal variable x and the vertical variable y.
Your hill is described by an equation of the form f(x) = -x.
Your rock's path is given by a parabola, which you should be able to calculate using the initial conditions you gave (x=y=0 initially, vy = 0, vx = 15 m/s). Let's call this trajectory g(x).
Now you can find the point at which the rock will hit the hill by equating f(x)=g(x). This will let you know far down the rock "fell" (remember, its motion in the y-direction is independent of its motion in the x-direction. It is merely a free fall). Knowing how far something falls, you can calculate the time it takes it to fall.

If you get stuck get come back & ask again :).

------
Assaf
http://www.physicallyincorrect.com/"

 

[physucsc11]

That's basically what I was doing. I realized that since it is a 45 deg angle the rock will travel the same distance vertically as horizontally before it hits the ground, so I used the equation y = -1/2gt^2 and plugged in d = vx*t for y. So I got vx*t = -1/2gt^2 , and solving for t I get t = (2vx +- sq.root( (2vx)^2))/2g). So the answer turns out to be 3.2 sec. However, the book gives 5.1 sec. Am I doing something wrong or is the book wrong?

 

[physucsc11]

So can anyone help?