# Projectile lands on an incline.

1. Jan 12, 2014

### moria

1. The problem statement, all variables and given/known data

My problem looks like this:
http://i.imgur.com/3M9wE43.png

I need to find the point of impact and the time the body traveled.

2. Relevant equations

3. The attempt at a solution

I tried to find the time for 25 meters and with that the vertical velocity and height.
Ended up getting: t at 25m = 1.96 s , h=16.83 m vi=-1.03 m/s
I tried this: hf=vx*sin37*t = 7.67 m/s*t , it should mean that for every seconds the slope rises 7.66m in height.
After that I set up: hf = h + vi*t - 1/2*g*t^2 and got:
7.67*t = 16.83 - 1.03*t - 4.905*t^2
-4.905*t^2 - 8.7*t + 16.83 = 0
4.905*t^2 + 8.7*t - 16.83 = 0

The positive t was 1.17 s

Total time = 1.96 s + 1.17 s = 3.13 s

Height:
h = vi*t - 1/2*g*t^2
h = 18.20 m/s * 3.13 s - 1/2 * 9.81 m/s^2 * (3.13s)^2
h = 8.91 m

distance = vx*t
distance = 12.75 m/s * 3.13 s = 39.9 m

My problem is that the height and distance doesn't match, the tangent of the angle is: height/distance = 8.91/39.9-25=8.91/14.9=0.5979 , with arctg I get around 31 degrees, shouldn't I get 37 degrees here? This is what bothers me. Is the method I used correct? If not, what is the method I should use to solve it?

Last edited: Jan 12, 2014
2. Jan 12, 2014

### SammyS

Staff Emeritus
Why use 3.13 sec. for the time to find the height?

3. Jan 12, 2014

### moria

Shouldn't the path of of the trajectory be the same as when I'm solving it without slope?
Wouldn't solving the height throught tg(37)*14.9 change the path?

Clear me up if I'm wrong, cause that is the part that bothers me the most about this problem.

4. Jan 12, 2014

### SammyS

Staff Emeritus
Yes, of course you're right about that.

You could use the equation for hf from the ramp, along with t = 1.17 s.

But looking at you're initial work, how did you get vx and vy ?

5. Jan 12, 2014

### barryj

Try the following. Assume the projectile lands on the ramp a distance d horizontally from the base(where the ramp starts) and a distance h higher than the base. d and h are related by the angle 37 degrees. Now do the standard projectile equations d(vertically) = v(0)t = (1/2)at^2 and d(horizontally) = vt. If I say much more I would work the problem.

6. Jan 12, 2014

### moria

Vy=sin(55)*v and Vx=cos(55)*v v is 22.22 m/s

What to do with the 25 meters? I know how to solve it when the projectile is fried from the bottom of the ramp.

tg(alpha)=Sy/Sx
tg(alpha)=(vy*t-gt²/2)/(vx*t) and so on, but where do the 25m fit into all that?

7. Jan 12, 2014

### SammyS

Staff Emeritus
moria,

It looks like you should use tan(37°), not sin(37°) for the ramp, hf.

Otherwise, your method appears to give a good result.

Last edited: Jan 12, 2014
8. Jan 12, 2014

### barryj

From the initial velocity of the projectile, you calculate the vertical, Vv, and horizontal velocity,Vh, knowing the angle of course.
The horizontal equation would be 25 + d = Vh * t
Now write the vertical equation. Remember that d and h are related by the angle of 37 deg.

9. Jan 12, 2014

### moria

With the risk that I might write bullsh*t:
25+d=vx*t
d=vx*t-25

tg(37)=h/vx*t-25
h=tg(37)*(vx*t-25)

h=vy*t-4.05t²
tg(37)*(vx*t-25)=vy*t-4.905t²

Like this?

10. Jan 12, 2014

### barryj

d=vx*t-25 correct
h=vy*t-4.05t² correct

tg(37)*(vx*t-25)=vy*t-4.905t² now solve for t and you are basically done. Yes, its a quadratic.

No

11. Jan 12, 2014

### moria

You are right, when I do that I get 1.06 seconds and when I add it to 1.96 I get 3.02 seconds.

And doing this gets me the same result, 3.02 seconds.

Thanks for the help.