# Boulder rolling off of cliff: will it hit the village?

1. Sep 4, 2015

### Entr0py

1. The problem statement, all variables and given/known data
A large boulder rests on a cliff 400 m above a small village in such a position that if it should roll off, it would leave with a speed of 50 m/s at an angle of 30 degrees below the horizontal. There is a pond with a diameter of 200 m, with its edge 100 m from the base of the cliff. The village houses are at the edge of the pond. A physics student says that the boulder will land in the pond. Is he right?

2. Relevant equations
y= y0 + v0yt + 1/2gt^2

x=R=v0xt

3. The attempt at a solution
Firstly, I need to find the vector components of the velocity. Since theta is -30 degrees and the magnitude of the velocity is 50 m/s, I find that v0x=50cos(30) which is 43.3 m/s and v0y=-50sin(30) which is -25 m/s.

Next I want to find out how long it would take the boulder to fall a vertical distance of 400 m (in this case, -400 m since i chose my y axis to be positive in upwards direction). I use the equation y= y0 + v0yt + 1/2gt^2. I take y0 to be 0 m and y to be -400. Using the quadratic formula, I find that t=6.84 s.

Then I want to find the range or the horizontal distance the boulder travels, which I find using the equation x=v0xt. Since v0x=43.3 m/s and t=6.84 s, I plug those in to get x=296 m.

I say that the physics student is right, that the boulder will land in the pond.

Last edited: Sep 4, 2015
2. Sep 4, 2015

### haruspex

How do you get a launch angle of 30 degrees?

3. Sep 4, 2015

### Entr0py

From my texbook. The author puts in the diagram that the angle below the horizontal is 30 degrees.

4. Sep 4, 2015

### Entr0py

Would that mean that v0y=-(50sin(30) or -25 m/s?

5. Sep 4, 2015

### haruspex

Yes. I was just about to point that out.

6. Sep 4, 2015

### Entr0py

Then if I plug that in the quadratic formula, t=6.84 s. Now if I plug that into eqn. to find range I get 296.2 m. After all the physics student was right and the boulder WOULD NOT hit the village.

7. Sep 4, 2015

### haruspex

Looks right to me.

8. Sep 4, 2015

### Entr0py

Yes! I was looking at this question last night, thinking "I can't wait to solve this!" Next time, I'll be sure to look at the launch angle very carefully. Thanks for the response.

9. Sep 4, 2015

### Entr0py

EDIT: the physics student is right.

10. Sep 4, 2015

### haruspex

... and perhaps mention it in the first post? :-)

11. Sep 4, 2015

### Entr0py

Yes, I will. I just looked at the diagram and said "Look at that 30 degrees!" but didn't notice that it was BELOW the horizontal until I rechecked.

12. Sep 4, 2015

### haruspex

Had it been above the horizontal it no doubt would have stayed safely in place on the cliff.

13. Sep 4, 2015

### Entr0py

That didn't even cross my mind. I need a common sense pill.

14. Sep 4, 2015

### Entr0py

Now to find how fast the boulder is going when it hits the ground I would use vf^2=v0^2 + 2a(x-x0). Which would mean that vf^2=50^2 - (19.6*-400) or vf=-101.7 m/s?

15. Sep 4, 2015

### Entr0py

Also, wouldn't its horizontal component of velocity just before it hits be found by v(x)= v0x +a(x)t? But since there is no horizontal acceleration, then it would be v(x)=v0x, since gravity does not act horizontally on the boulder, which would make v(x)=50cos(-30) or 43.3 m/s.

16. Sep 4, 2015

### haruspex

Yes.
Yes.

17. Sep 4, 2015

Gracias