Gravity's opposition to accleration

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I'm creating a pneumatic launching device to shoot projectiles vertically. I guess the closest thing to compare it to would be a regular gun. Once it leaves the barrel, it will be in a constant state of deceleration until it reaches its peak (no additional thrust provided).

Ignoring frictional losses, is there a systematic way to calculate the projectiles rate of deceleration? Does the weight matter (i.e. will a tennis ball and a bowling ball decelerate at the same rate)? Does the exit speed matter?

Ideally I would like to take a projectile, knowing the force provided by the pneumatic actuator and the weight of the projectile, calculate exactly how high the projectile will go. Then once I've mastered calculating it in a frictionless environment, put it in a wind tunnel and calculate losses due to friction.

Also, just a general question somewhat related, in an environment with friction, will it take the exact same amount of time to reach ground-to-peak as peak-to-ground?
 

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  • #2
DaveC426913
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Ignoring frictional losses, is there a systematic way to calculate the projectiles rate of deceleration? Does the weight matter (i.e. will a tennis ball and a bowling ball decelerate at the same rate)? Does the exit speed matter?
Yes - they will accelerate downward at 9.8m/s^2.

No weight doens't matter. Yes, the tennis ball and hte bowling ball will fall at the same rate - as per Galileo's experiment in the Leaning Tower.

No, not for acceleration.

Also, just a general question somewhat related, in an environment with friction, will it take the exact same amount of time to reach ground-to-peak as peak-to-ground?
With friction? No. It will take longer on the way down.
 
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  • #3
ranger
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Ideally I would like to take a projectile, knowing the force provided by the pneumatic actuator and the weight of the projectile, calculate exactly how high the projectile will go. Then once I've mastered calculating it in a frictionless environment, put it in a wind tunnel and calculate losses due to friction.
You will not need the weight of the projectile, but the mass. Here is a simulator:
http://galileo.phys.virginia.edu/classes/109N/more_stuff/Applets/ProjectileMotion/jarapplet.html
In addition, I think you need to read up on projectile motion to get a grasp on things such as maximum height and so on.
http://id.mind.net/~zona/mstm/physics/mechanics/curvedMotion/projectileMotion/generalSolution/generalSolution.html [Broken]
 
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  • #4
russ_watters
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Well, that calculator used mass and that's fine since mass and weight are proportional (close to earth), but really since drag is a force, you need weight. Here's the drag equation: http://www.grc.nasa.gov/WWW/K-12/airplane/drageq.html -- just add a +w on the end of it. And heck, for this limited purpose, you can even make your own psuedo-Cd by dropping all the other variables into it for simplicity, ending up with: D = Cd*V^2 + w

[edit] You can always keep the two acceleration terms separate until you get to the acceleration equation (which is what you are really looking for) and just say a = D/m + 9.8 (is that what you were getting at, ranger?)

Also, if you are going to be measuring drag in a wind tunnel, then life is easy - using the drag equation, you can easily calculate the drag coefficient and plug it back into a spreadsheet for numeric integration of the trajectory.

Also, in an impulse situation, force can be a tricky thing to use. If you can make it this simple, you can start with a pressure and initial and final volume of gas, calculate the pressure when the ball exits, take the average and multiply by area to get force. Then with force and distance, you can find the acceleration/exit velocity. These things can get iffy in an impulse situation, though, depending on the particulars of the canon.
 
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Thanks for all the info! I haven't studied physics since high school and it's difficult getting all the information you provided if you don't know where to start. It looks like NASA has a whole slew of information so I'll be very busy the next few days on their site.
 

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