How do you find the acceleration needed to clear a jump?

In summary, the student attempted to solve a homework problem involving a jump, but did not have the information needed to complete the task. He was helped by another user who shared equations that he could use to solve for the horizontal and vertical displacements of the trajectory.
  • #1

Homework Statement


e4091526.png


Homework Equations


3ba92745.png


The Attempt at a Solution


I used Pythagorean theorem to find the length of the ramp (25^2+18^2 = 949) and found the angle of elevation using tangent (Tanθ=18/25) but then got stuck on what formula to use.
 

Attachments

  • e4091526.png
    e4091526.png
    9 KB · Views: 838
  • 3ba92745.png
    3ba92745.png
    909 bytes · Views: 703
Physics news on Phys.org
  • #2
First, what velocity does it need when it leaves the ramp? Intuitively, you can see that if you leave too slowly, you're going to fall down into the gap. So what determines whether you'll make it?

The horizontal motion is constant velocity ##v \cos(\theta)##. That will determine how long it is in the air. Use the equation for d but with a = 0 since it's constant velocity. Horizontal d is fixed. You're solving for time.

The vertical motion is accelerated motion with initial vertical velocity ##v \sin(\theta)##. That will determine where the car is after that much time. Use the equation for d but with a = -9.8 m/s^2.
It will make the jump if that vertical displacement is enough. If d from where it left the ramp is enough to make it to the platform. How much higher is the platform than the end of the ramp? That's what the vertical d needs to be.

OK, now you know what the velocity needs to be at the end of the ramp. It needs to have enough acceleration to get to that velocity within the length of the ramp. Is there a formula that relates acceleration, distance and velocity? There is. Use that one, since you know two of the three variables and can solve for the other one (acceleration).
 
  • #3
So I know the length of the ramp and the angle but how exactly do I find the velocity? For part one you mentioned using v cos(θ) to determine how long the car will be in the air but how do I use that formula if I don't know what v (assuming that v means velocity) is?
 
  • #4
physicstryhard said:
I don't know what v (assuming that v means velocity) is?
So leave it as unknown. You can still write the equation relating it to time of flight. Likewise for the vertical motion. Two equations, two unknowns: solve.
 
  • #5
physicstryhard said:

Homework Statement


View attachment 230985

Homework Equations


View attachment 230986

The Attempt at a Solution


I used Pythagorean theorem to find the length of the ramp (25^2+18^2 = 949) and found the angle of elevation using tangent (Tanθ=18/25) but then got stuck on what formula to use.

It seems to me that you need to know the distance across the gap. Do you have that information?
 
  • Like
Likes CWatters
  • #6
PeroK said:
It seems to me that you need to know the distance across the gap. Do you have that information?
nope. it would be much easier if I did but the assignment did not provide it.
 
  • #7
haruspex said:
So leave it as unknown. You can still write the equation relating it to time of flight. Likewise for the vertical motion. Two equations, two unknowns: solve.

So by leaving it as unknown that gives me an equation: 25=vi*t + 0. Now I don't know the time or the velocity how would I solve this?
 
  • #8
Suppose the gap was 5000 km. You would have to go very fast to make the jump. I also believe it's necessary to solve the problem.
 
  • #9
physicstryhard said:
So by leaving it as unknown that gives me an equation: 25=vi*t + 0. Now I don't know the time or the velocity how would I solve this?
Wrong equation. How is the 25m relevant here? We want the launch speed at the top of the ramp, and the 25m is behind.
As others have noted, the question cannot be solved without knowing the width of the gap. E.g. with a gap of 2m it is impossible; no matter how fast you go you won't make it.
Let the gap width be x. If the time of flight is t and the launch speed v, what equations can you write for the horizontal and vertical displacements of the trajectory?
 
  • #10
I see, alright thanks for your help. I it makes sense that this would be impossible since the width of the gap isn't stated so I suppose it could be any number like 5000 km or something ha ha.
 
  • #11
physicstryhard said:
I see, alright thanks for your help. I it makes sense that this would be impossible since the width of the gap isn't stated so I suppose it could be any number like 5000 km or something ha ha.
There is nothing to stop you following my advice in post #9. Just let the gap width be unknown x and find an answer in terms of that.
 

1. What is acceleration?

Acceleration is the rate of change of velocity over time. In other words, it is how quickly an object's velocity changes.

2. How is acceleration related to clearing a jump?

In order for an object, such as a person or a vehicle, to clear a jump, it needs to have a certain amount of upward acceleration to counteract the force of gravity pulling it down.

3. How do you calculate the acceleration needed to clear a jump?

The acceleration needed to clear a jump can be calculated using the formula a = (vf^2-vi^2)/2d, where a is acceleration, vf is final velocity, vi is initial velocity, and d is the distance of the jump. This formula takes into account the starting and ending velocities as well as the distance traveled.

4. What other factors can affect the acceleration needed to clear a jump?

The acceleration needed to clear a jump can also be affected by the weight and mass of the object, as well as any air resistance or friction present during the jump.

5. How can acceleration be controlled to successfully clear a jump?

In order to control the acceleration needed to clear a jump, factors such as the angle of the jump, the speed and force applied, and the trajectory of the jump can all be adjusted. Additionally, having a good understanding of physics and the principles of acceleration can help in effectively controlling the jump.

Back
Top