# Homework Help: What is the value of this distance up the slope?

1. Sep 15, 2004

### cdhotfire

I got this problem for physics and i cant seem to get it, it goes like:

A projectile is launched at an angel of 40 degrees with an initial velocity of 100 m/s. One hundred meters away is the beginning of a hill that slopes upward at an angle of 20 degrees. The projectile strikes the hill a distance of L up the slope. What is the value of this distance up the slope?

Okay so far ive gotten:
H | a=0 | Vi=76.6 | x=100 | t= 1.31 s
V | a=-9.81 | Vi=64.28 | x=75.79 |

thats all ive gotten, but i cant figure out how to c where it lands. Maybe someone can give me a hint or sometin im really stuck.

Thanks You.

2. Sep 15, 2004

### Pyrrhus

The tangent of 20 degrees is the slope of the straight line, so y = tan(20)x +b, also when Y of the Parabole Trajectory is equal to the Y of the straight line, they collide. In my opinion you could find the trajectory equation for your parabole and solve for x to find the x value it will have when Y of the parabole is equal to Y of the straight line, then you can find the exact coordinates (X,Y) for the collision, then you probably can apply the distance formula of two points for (100,0) and (X,Y).

Last edited: Sep 15, 2004
3. Sep 15, 2004

### cdhotfire

hmm, good idea ill try that out.
thxs.

4. Sep 15, 2004

### cdhotfire

one more thing, how do you know that the slope of the line is tan 20*?

5. Sep 15, 2004

### Pyrrhus

definition of slope, m = y/x, tangent of the angle is equal to y/x.

6. Sep 15, 2004

### cdhotfire

wow ur right, hehe, thxs.

7. Sep 15, 2004

### cdhotfire

Hmm, im sorry to bother, but i dont know how to find equation for parabola.

8. Sep 15, 2004

### Pyrrhus

Well are you familiar with parametric equations?

9. Sep 15, 2004

### cdhotfire

No im not

10. Sep 15, 2004

### Pyrrhus

Well, it doesn't matter, it was just easier for me to point you the right direction but it's ok.

Let's see we know this equation from kinematics with constant acceleration:
(Looks like LateX isn't working, oh well, i'll try to write it as clear as possible.)

X = Xo + Vot + (1/2)at^2

Let's start analyzing horizontal trajectory for this case.

x = VoCos(angle)t

Now let's do the same for vertical trajectory for this case.

Y = VoSin(angle)t - (1/2)gt^2

If we want to get our Y = f(x), then we need to eliminate, get rid of t in both of our equations.

so t = x/(VoCos(angle))

substuting it in our second equation we will have

Y = VoSin(angle)*x/(VoCos(angle)) - (1/2)g*(x/(VoCos(angle)))^2

Y = tan(angle)x - (g*x^2)/(2*Vo^2*Cos^2(angle))

Now we have our Y= f(x), for any value of x we can get a value of y, you can notice it's a parabole because of the x^2.

Last edited: Sep 15, 2004
11. Sep 15, 2004

### cdhotfire

says latex is invalid.

12. Sep 15, 2004

### Pyrrhus

Yes, give me a minute i'm editing my message with the info you need.

13. Sep 15, 2004

### cdhotfire

thxs

14. Sep 15, 2004

### Pyrrhus

See if you can understand what i said.

15. Sep 15, 2004

### cdhotfire

what does the a and g represent. These formulas dont seem familiar.

16. Sep 15, 2004

### Pyrrhus

a represents acceleration and g is for gravity.

It's a kinematic equation...

X = Xo + Vot + (1/2)at^2
X - Xo = Vot + (1/2)at^2

sometimes teachers put it as
d= X - Xo
d = Vot + (1/2)at^2

17. Sep 15, 2004

### cdhotfire

oops i meant to say o not a .

18. Sep 15, 2004

### Pyrrhus

It's a notation used that means initial, so Vo will be initial speed.

19. Sep 15, 2004

### cdhotfire

Plus i dont think the teacher has teached us kinematics, all weve learned so far are these formulas:
x=Vi t + 1/2 a ^2
a=(Vf - Vi)/ t
Vavg= (Vf + Vi)/2= x/t
Vf^2 - Vi^2= 2 a x

20. Sep 15, 2004

### cdhotfire

wait wait its all starting to make sense now, i think ive got it now. Thank you very much for all the help, I appreciate it. Take it ez.