# Angle that the stone is thrown off of the cliff?

1. Jun 4, 2009

### balling12

1. The problem statement, all variables and given/known data
A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 16.0 m/s. The cliff is h = 62.0 m above a flat horizontal beach.

How long after being released does the stone strike the beach below the cliff?

With what speed and angle of impact does the stone land?

How do I solve without knowing the angle that the stone is thrown off of the cliff?

2. Jun 4, 2009

### Cyosis

Re: help

The angle is given by the bolded part.

3. Jun 4, 2009

### balling12

Re: help

so the angle is 0 or 180?

4. Jun 4, 2009

### Cyosis

Re: help

Technically zero, but what do you know about the components if an object is thrown horizontally? You don't need to do any trigonometry.

5. Jun 4, 2009

### balling12

Re: help

I tried working it without using trig but I got the wrong answers.

6. Jun 4, 2009

### Cyosis

Re: help

Show your steps so we can have a look at it.

7. Jun 4, 2009

### balling12

Re: help

Ok so you said I didnt need to use trig so to get the time i simply divided 62 by 9.8 and got 6.3 seconds.

8. Jun 4, 2009

### dx

Re: help

And why do you think that would give you the time?

9. Jun 4, 2009

### Cyosis

Re: help

That is not right, 62 meters divided by 9.8 m/s^2 gives 6.3 s^2. Do you know of any kinematic equations that give you displacement as a function of time for a uniformly accelerating object?

10. Jun 4, 2009

### balling12

Re: help

Because i figured if the height is 62m and the gravitational constant is 9.8m/s dividing would give me the time of the fall.

11. Jun 4, 2009

### Cyosis

Re: help

Well if you had just looked at the dimensions you would have figured it was wrong. As I said your calculation yields 6.3 seconds squared, which is not the unit of time.

Edit: the gravitational constant is 9.8 m/s^2 not 9.8m/s.

12. Jun 4, 2009

### dx

Re: help

So you simply guessed? 9.8 m/s² is the acceleration due to gravity. Do you recognize this equation: s = ut + ½at² ?

13. Jun 4, 2009

### balling12

Re: help

yes i recognize that equation

14. Jun 4, 2009

### Cyosis

Re: help

What are s, u and a in this problem?

15. Jun 4, 2009

### balling12

Re: help

s=62 u=? a=-9.8m/s
I tried using this equation at first but ran into the same problem because i was unsure of what u is.

16. Jun 4, 2009

### dx

Re: help

That's the equation to use here. This is what it means: if a particle startes out with speed u (in some direction), and has an acceleration a(in that direction), then the distance travelled by the particle after time t (in that direction) is s = ut + ½at².

17. Jun 4, 2009

### Cyosis

Re: help

You should know what u is after reading dx's post. One thing though if you set s=62 then a needs to be positive or if you take a to be negative then s=-62. Also I generally use this notation:

$s=\frac{1}{2}at^2+v_0t+s_0$

The v_0 symbol is slightly more suggestive than the u symbol.

18. Jun 4, 2009

### balling12

Re: help

so would u=16?

19. Jun 4, 2009

### Cyosis

Re: help

In which direction is gravity accelerating the stone? In which direction is the stone moving with 16m/s?

20. Jun 4, 2009

### balling12

Re: help

actually horizontally