# Min. Speed of a Rock

1. Nov 20, 2007

### surfhare75

Here is the Problem. A 2.8-kg block slides over the smooth, icy hill shown in the figure .The top of the hill is horizontal and 70 m higher than its base. i attached the Figure Below.

What minimum speed must the block have at the base of the hill so that it will pass over the pit at the far side of the hill without falling into it?

I tried using .5mv1^2+m1gH=.5mv2^2+m2gH but I need to know the gap between the pit which is 40m. I believe I use H=.5gt^2 to find the time and use projectile motion x=vosin(x)t to find the distance.

#### Attached Files:

• ###### yf_Figure_7_35.jpg
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2. Nov 20, 2007

### Bill Foster

I can't see the picture.

Is the hill curved?

3. Nov 20, 2007

### surfhare75

The picture should be loaded now, but in the picture the hill is curved but does not seem to be stated specially. This is from a chapter on potential and conservation energy

4. Nov 20, 2007

### Bill Foster

Cant' see it. It says: "Attachments Pending Approval"

5. Nov 20, 2007

### azatkgz

I also cannot see figure.

6. Nov 20, 2007

### azatkgz

I think if the top of the hill is horizontal,then we can use

$$x=v_{top}t$$

$$y=\frac{1}{2}gt^2$$

7. Feb 16, 2009

### james brug

The solution to this should be the initial velocity it takes to get to top of the hill plus the velocity needed to clear the pit, right? I get about 56.8 m/s which is apparently wrong.

$$\sqrt{1372}\;m/s+19.8\;m/s$$

8. Feb 16, 2009

### james brug

9. Feb 16, 2009

### james brug

10. Feb 16, 2009

### Duncan1382

I'm not sure if this is right, but this is what I came up with:

1) I found the speed that the object needed to be going to clear the jump.. not too hard. I assumed the ground was horizontal at the launch angle.
2) I used Energy to solve the problem.

Speed to Clear:
$$\Delta$$y = .5at2
-20m = -4.9$$m/s$$2t2
t = 2.02 seconds

no wind resistance = no acceleration

v = delta x / t = 40m / 2.02s = 19.80m/s

--
Etop = Ebottom

2.8kg(g)(70m) + .5(2.8kg)(19.8$$m/s$$)2 = .5 (2.8kg)v2
v = 42.00$$m/s$$

11. Feb 17, 2009

### james brug

Yes, that is correct.

$$\sqrt{1372\;m^2/s^2+(19.8\;m/s)^2}$$= Life, the Universe, and Everything