# Motion in 2 dimentions

1. Feb 13, 2004

### Beretta

A worker on the roof of a house drops her hammer which slides down the roof at constant speed of 4m/s. Th roof makes an angle of 30 with the horizontal, and the lowest point is 10m from the ground. what is the the horizontal distance traveled by the hammer between the time is leaves the roof of the house and the time it hits the ground?

What happen here is that I assumed that theta is 60 since it makes an angle if 30 with the horizontal and since the object is sliding down. Thus it leaves the roof on a 60 degree angle.

vx = 4m/s cos 60 = 2m/s
vy = 4m/s sin 60 = 3.46m/s

x(t) = x0 + vxt

y(t) = y0 + v0t - (1/2)g(t^2)
10m - 3.46m/s(t) + (0.5)(9.81m/s^2)t^2 = 0

I am ending up with a negative root Delta.
May you help me pls?

2. Feb 13, 2004

### Michael D. Sewell

Motion in 2 dimensions

try:
vx = 4 cos 30
vy = 4 sin 30

3. Feb 13, 2004

### Beretta

Why would the angle be 30 degree not 60 degree?

4. Feb 13, 2004

### Chi Meson

What you want is the angle that the hammer is moving with respect to the horizontal. THe hammer travels the same plane as the roof, which is sloped 30 degrees to the horizontal. THe 60 degrees is the angle made with a vertical line. IF you use the angle from vertical, you have to switch your sines and cosines.

You need to find the time of travel using your vertical information. It will send you into a quadratic solution. If you are getting a negative under the radical, it's probably because you are not recognizing that both acceleration and y-displacement are negative values (since you are calling initial y-velocity negative).

5. Feb 13, 2004

### Beretta

Should I leave it 30 then? I am really confused!

6. Feb 13, 2004

### Moose352

Try drawing it out. You can use either 30 or 60, but you must remember to use the right trignometric function to find the appropriate component.

7. Feb 13, 2004

### Michael D. Sewell

Motion in two dimensions

Chi Meson and Moose are right. You should start this out by drawing the problem(always!). Watch your signs, you are in charge so you get to decide which direction is negative, however, you also have to answer to your instructor.
The important thing to be learned from this problem is that the speed is proportional to the length of the sides of the right triangle. Please remember this, think about it, and use it often so that you don't forget it.
Trigonometry can be very confusing at first, but if you bear this one thing in mind it will help you greatly; c^2 = a^2 + b^2.
Everything else in trigonometry is just a restatement of this.
To solve your problem you could use:
vx = 4 cos 30
vy = 4 sin 30
or
vx = 4 sin 60
vy = 4 cos 60
The result will be the same either way. It's not just a good idea, it's the law.

8. Feb 13, 2004

### Beretta

I am really greatful to everyone. Thank you all. One more question though, shouldn't be vx = -4 cos 30 and vy = -4 sin 30 since the object is sliding in the negative direction?

9. Feb 13, 2004

### Moose352

Anything you want, as long as you are consistent. Generally, left negative, right positive is the traditional system.