# Projectile problem--A rock is kicked off a 45 Degree hill at 15m/s.

by Kizaru
Tags: 15m or s, degree, hill, kicked, problema, projectile, rock
 P: 45 Hello. I tried this problem and it's the last one for my problem set due Thursday, but my answer differs from the given answer in the back of the book. After trying the solution again, I searched Google and found this. I'm having the exact problem. 1. The problem statement, all variables and given/known data A rock is kicked horizontally at 15m/s from a hill with a 45 degree slope. How long does it take for the rock to hit the ground? Vix = 15m/s Viy = 0 Yi = 0 Xi = 0 2. Relevant equations Vx = Vix X = Xi + Vxo*t Vy = Viy - gt Y = Yi + Viy*t - (1/2)g*t^2 3. The attempt at a solution I tried equating the X and Y position functions, but this only provides the intersection between them--it does not provide the intersection between the rock and the hill. My second method is identical to the one described in the link mentioned earlier in my post. The problem is that my answer isn't the same as the one in the back of the book, and I'm not sure if it's a problem on my end. I don't know what to do after this. Any help would be great :)
 PF Patron Sci Advisor Emeritus P: 4,974 If the rock lands at the same height it was launched one can obtain the time from the y position equation since y=0.
P: 45
 Quote by Kurdt If the rock lands at the same height it was launched one can obtain the time from the y position equation since y=0.
The rock will never pass that point because it's launched horizontally (no initial Y velocity) off a 45 degree hill (the hill is declining). The only time the rock is at y=0 is when t=0.

Imagine graphing y=-x and y=-x^2. If you focus on only the third quadrant, you have what looks like the hill (y=-x) and the path of the rock (y=-x^2), but it's not that easy to find an equation that represents the path of the rock.

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