Calculation of Ramp Jump Distance Using Projectile Motion Formulas?

  • #1
Amaretto
5
0
Homework Statement
Two friends build a ramp with a 10° angle at both Sides of the shore in between is a creek. The travel with a Velocity of 7m/s where the acceleration is capped so they always drive 7m/s. What distance do they travel when they drive over the ramp? No Air resistance Nor friction. My Language is bad, I'm not native
Relevant Equations
g=9.81m/s^2
x(t)=x0+v0*t+1/2*a*t^2
b=10°
I created a Vector which describes the direction of the medium. with r=(x(t),y(t))=((x0+cos(b)*t+0*t^2/2),(y0+sin(b)*t-9.81*t^2/2))
I created a vector for the acceleration a=(0,-9.81)
Now only the y(t) is relevant since x(t)
acceleration is zero. sin,cos describe the direction of the velocity without acceleration.
Now i wanted to set y0=0 and find where y=0. Because i think thats the moment he lands. I solved 0=0+t*sin(10°)-9.81t^2/2 to find t and got 0.0355s put that into s=v*t and got like 25cm. I think there`s something wrong. What do yall think?
 
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  • #2
Amaretto said:
Homework Statement: Two friends build a ramp with a 10° angle at both Sides of the shore in between is a creek. The travel with a Velocity of 7m/s where the acceleration is capped so they always drive 7m/s. What distance do they travel when they drive over the ramp? No Air resistance Nor friction. My Language is bad, I'm not native
Relevant Equations: g=9.81m/s^2
x(t)=x0+v0*t+1/2*a*t^2
b=10°

I created a Vector which describes the direction of the medium. with r=(x(t),y(t))=((x0+cos(b)*t+0*t^2/2),(y0+sin(b)*t-9.81*t^2/2))
I created a vector for the acceleration a=(0,-9.81)
Now only the y(t) is relevant since x(t)
acceleration is zero. sin,cos describe the direction of the velocity without acceleration.
Now i wanted to set y0=0 and find where y=0. Because i think thats the moment he lands. I solved 0=0+t*sin(10°)-9.81t^2/2 to find t and got 0.0355s put that into s=v*t and got like 25cm. I think there`s something wrong. What do yall think?
Do you have a picture that accompanies this problem?
 
  • #3
The time you found is incorrect. You need to find the correct solution to the quadratic.
 
  • #4
kuruman said:
The time you found is incorrect. You need to find the correct solution to the quadratic.
I can't see what is being asked? There is no height to the ramps, there is no distance across the stream to the other ramp?
 
  • #5
kuruman said:
The time you found is incorrect. You need to find the correct solution to the quadratic.
So the equation is the problem. But I don't know where the mistake is or is the approach wrong. I'm sure I solved it properly. It was t= 2*sin(10°)/9.81
 
  • #6
erobz said:
I can't see what is being asked? There is no height to the ramps, there is no distance across the stream to the other ramp?
I was stuned about that either but I dont think thats important here. I think it is more so about calculating with vectors
 
  • #7
Amaretto said:
So the equation is the problem. But I don't know where the mistake is or is the approach wrong. I'm sure I solved it properly. It was t= 2*sin(10°)/9.81
And where is the velocity of 7 m/s?
 
  • #8
kuruman said:
And where is the velocity of 7 m/s?
Now that I notice.. . Do i put it infront of sin?
 
  • #9
Amaretto said:
I was stuned about that either but I dont think thats important here. I think it is more so about calculating with vectors
If it's a ramp takeoff to a ramp landing across a stream gap this stuff matters...Unless the ramps are identical and they are asking what is the maximum gap such that they make the jump?
 
  • #10
erobz said:
I can't see what is being asked? There is no height to the ramps, there is no distance across the stream to the other ramp?
All that is true. However, the OP seems to be concerned with the horizontal distance they travel when they return to the same height. When they do so, it is apparently immaterial to the OP whether there is terra firma under them or an abyss.
 
  • Wow
Likes erobz
  • #11
Amaretto said:
Now that I notice.. . Do i put it infront of sin?
What do you think? You said you want to do it by vector addition. What vectors are you adding?
 
  • #12
kuruman said:
What do you think? You said you want to do it by vector addition. What vectors are you adding?
I just put it infront of sin, since v0 is the starting velocity. I think 1.7m make more sense now. Thank you
 
  • #13
GolferTriangle.png
The vectors that you add are ##~\mathbf v_0~t_{\!f}~## in the initial direction of motion and ##~ \frac{1}{2}\mathbf g~t_{\!f}^2~## straight down. Here, ##t_{\!f}## is the time of flight. These two vectors add to give the final position vector ##\mathbf L## as shown in the figure on the right.
 

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