Attempting to calculate maximum potential distance of a projectile from set height

  • Thread starter shini
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  • #1
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Was not sure where to post this specifically, but I am looking to calculate the possible distance traveled by by a projectile from a set height, speed, and release. Its just curiosity from seeing a new ride being brought to my local theme park.

Height of release is 121.92 meters.
Mass of projectile is 90.72 kg.
Radius is 6.01 meters.
Velocity is 15.65 m/s.
Centripetal force is 3697.07 N.
Angle of release is 45°.

Thanks for any input on this.
 

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  • #2
berkeman
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Was not sure where to post this specifically, but I am looking to calculate the possible distance traveled by by a projectile from a set height, speed, and release. Its just curiosity from seeing a new ride being brought to my local theme park.

Height of release is 121.92 meters.
Mass of projectile is 90.72 kg.
Radius is 6.01 meters.
Velocity is 15.65 m/s.
Centripetal force is 3697.07 N.
Angle of release is 45°.

Thanks for any input on this.

Welcome to the PF.

How did you get that centripital force? Do you have any pics of the ride?
 
  • #3
35,392
11,744


Neglecting air resistance (a good idea if you do not want to simulate it), you do not need any parameters apart from the initial velocity and the angle. The device to accelerate the object does not matter.

Can you calculate the initial vertical and horizontal velocity?
Based on this, can you calculate the time at which the vertical velocity gets 0?
Can you calculate the height at this point?
Can you calculate the time to fall down, if you know this height?
If you know the total flight time, can you calculate the distance?
That should help at the steps.
 
  • #4
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got the centripetal force from f= mv2[itex]/[/itex]r

forgot to mention its a swing style ride

I am assuming a perfect release angle of 45 degrees... just to try and simplify...
 

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  • #5
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mfb, you were assuming initial and final height were to be the same, which they are not. so you do need the initial height as well
[edit: also, i am a bit surprised you would calculate the time to go up and the time to go back down separately... but i guess it is a good excercise for the reader to see if they notice the symmetry :p then you completely do not need to calculate the height of the peak as well]

shini, centripetal force is irrelevant - it does not act after you have released the projectile.

angle of release may also not be 45 degrees. seems it actually is 0 degrees, i.e. moving entirely horizontally at release. it is not optimal for maximum distance, but i don't imagine you being able to impart any reasonable vertical component to the projectile velocity, so that is that. this actually simplifies the calculations a good bit.
 
  • #6
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Looking at this ride, I see that an experimentalist could make the front page of both reddit and youtube by determining the result in the manner [STRIKE]prefered by[/STRIKE] unique to their profession.
 
Last edited:
  • #7
35,392
11,744


mfb, you were assuming initial and final height were to be the same, which they are not. so you do need the initial height as well
No I were not.
[edit: also, i am a bit surprised you would calculate the time to go up and the time to go back down separately... but i guess it is a good excercise for the reader to see if they notice the symmetry :p then you completely do not need to calculate the height of the peak as well]
Do the calculation yourself, and you will see that this way avoids quadratic equations (it just requires to solve an equation of the type x^2=c for x).

angle of release may also not be 45 degrees. seems it actually is 0 degrees, i.e. moving entirely horizontally at release. it is not optimal for maximum distance, but i don't imagine you being able to impart any reasonable vertical component to the projectile velocity, so that is that. this actually simplifies the calculations a good bit.
I agree.
While 0° is not optimal*, it is quite close to that.


*looking at the context, I would think that "optimal" is more like "worst case" here.
 

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