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Projectile lands on an incline.

  • Thread starter moria
  • Start date
  • #1
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Homework Statement



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.

Homework Equations



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:

Answers and Replies

  • #2
SammyS
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Homework Statement



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.

Homework Equations



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

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?
 
  • #3
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0
Why use 3.13 sec. for the time to find the height?
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
SammyS
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Shouldn't the path of of the trajectory be the same as when I'm solving it without slope?
Wouldn't solving the height thought 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.
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
747
36
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
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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 ?
Vy=sin(55)*v and Vx=cos(55)*v v is 22.22 m/s

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.
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
SammyS
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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:
  • #8
747
36
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
5
0
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
747
36
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
5
0
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.
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.

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

Thanks for the help.
 

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