What is the value of this distance up the slope?

[cdhotfire]

I got this problem for physics and i can't 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 I've 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 I've gotten, but i can't figure out how to c where it lands. Maybe someone can give me a hint or sometin I am really stuck.

Thanks You.

 

[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).

 

[cdhotfire]

hmm, good idea ill try that out.
thxs.

 

[cdhotfire]

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

 

[Pyrrhus]

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

 

[cdhotfire]

wow ur right, hehe, thxs.

 

[cdhotfire]

Hmm, I am sorry to bother, but i don't know how to find equation for parabola.

 

[Pyrrhus]

Well are you familiar with parametric equations?

 

[cdhotfire]

No I am not

 

[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 + volt + (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.

 

[cdhotfire]

says latex is invalid.

 

[Pyrrhus]

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

 

[cdhotfire]

thxs

 

[Pyrrhus]

See if you can understand what i said.

 

[cdhotfire]

what does the a and g represent. These formulas don't seem familiar.

 

[Pyrrhus]

what does the a and g represent. These formulas don't seem familiar.

a represents acceleration and g is for gravity.

It's a kinematic equation...

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

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

 

[cdhotfire]

oops i meant to say o not a .

 

[Pyrrhus]

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

 

[cdhotfire]

Plus i don't think the teacher has teached us kinematics, all we've 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

 

[cdhotfire]

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

 

[Pyrrhus]

Those are the kinematic equations a little simplified.

 

[Pyrrhus]

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

No problem, Have fun solving it!

 

[cdhotfire]

hehe there will be tons of that.

 

[cdhotfire]

i got 104.37, does that sound right?

 

[Pyrrhus]

i got my (X,Y) as (637.40, 195.60), so my distance will differ using the Plane Distance Formula, SQRT((x-xo)^2 + (y-yo)^2). I got 571.89 meters.

 

[cdhotfire]

how u get that point i got (4.36, 1.57) and i did everything u said.

 

[Pyrrhus]

well first i calculated b, because we know nothing of b, remember the equation for a straight line is y = mx +b, so we know the slope is tan(20), and in our equation it should be y = tan(20)x + b, now we know a point our straight line goes through which is (100,0) so to find b we do, 0 = tan(20)*100 + b, so b = -tan(20)*100. Now we know all the values of our straight line equation, so we proceed to equal both the equations and solve for x.

tan(20)x -tan(20)*100 = tan(40)x - (9.8*x^2)/(2*100^2*Cos^2(40))

0 =tan(20)*100 + x(tan(40)-tan(20)) - (9.8*x^2)/(2*100^2*Cos^2(40))

0 = C + xB + Ax^2

C = tan(20)*100
B = (tan(40)-tan(20))
A = - (9.8*x^2)/(2*100^2*Cos^2(40))

I took the positive value of X of course.

I got X = 637.40, i plug it in both of the equations and got the same Y=195.60, then i used the formula above to calculate the distance.

 

[cdhotfire]

i c now i didnt get the b so then the whole thing went to chambles.
thxs