Calculation of Ramp Jump Distance Using Projectile Motion Formulas?

In summary, the calculation of ramp jump distance using projectile motion formulas involves determining the initial velocity, launch angle, and the effects of gravity. By applying the equations of motion, specifically the horizontal and vertical components, one can calculate the maximum distance a projectile will travel when launched from a ramp. Key factors include the angle of the ramp, the speed at which the object leaves the ramp, and the height from which it is launched, all of which influence the overall jump distance.
  • #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.
 

FAQ: Calculation of Ramp Jump Distance Using Projectile Motion Formulas?

What is the basic formula for calculating the distance of a projectile?

The basic formula for calculating the horizontal distance (range) of a projectile launched at an angle is given by: R = (v^2 * sin(2θ)) / g, where R is the range, v is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity (approximately 9.81 m/s²).

How does the launch angle affect the jump distance?

The launch angle significantly affects the jump distance. The optimal angle for maximum range in a vacuum is 45 degrees. Angles less than 45 degrees result in a shorter distance, while angles greater than 45 degrees also decrease range due to increased vertical motion, which reduces horizontal distance.

What role does initial velocity play in ramp jump distance?

Initial velocity is crucial in determining the jump distance. A higher initial velocity increases both the horizontal and vertical components of the motion, leading to greater overall distance. The relationship is quadratic, meaning that doubling the initial velocity will quadruple the range, assuming the angle remains constant.

How do you account for air resistance in ramp jump calculations?

Air resistance can be accounted for by using more complex equations that include drag force, which depends on the shape and size of the object, its velocity, and the density of the air. This requires numerical methods or simulations, as it complicates the simple projectile motion equations significantly.

What is the effect of height on ramp jumps?

The height from which a jump is made affects the initial vertical velocity and the time of flight. A higher launch point increases the jump distance because it allows the projectile to gain additional vertical distance before descending, thus increasing the overall range when combined with the horizontal motion.

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