How to calculate theoretical initial velocity?

[pill]

I can do with just an equation, but if anyone is willing to help:
Target is 25.6 m away from canon
Initial Velocity = 16 m/s
Time = 2.6 s
Range = 26.7m
No air resistance

 

[PeterO]

I can do with just an equation, but if anyone is willing to help:
Target is 25.6 m away from canon
Initial Velocity = 16 m/s
Time = 2.6 s
Range = 26.7m
No air resistance

Is your problem that you have been told the initial velocity is 16 and you don't know how it was calculated?

 

[genericusrnme]

I can do with just an equation, but if anyone is willing to help:
Target is 25.6 m away from canon
Initial Velocity = 16 m/s
Time = 2.6 s
Range = 26.7m
No air resistance

Initial Velocity = 16 m/s


If you mean final then you're going to have to tell me what this Range quantity is

 

[pill]

Initial Velocity = 16 m/s


If you mean final then you're going to have to tell me what this Range quantity is

To be honest, I'm not too sure. My instructions state, "Calculate the theoretical initial velocity (no air resistance) required to impact a target with the distance you used in the simulation." I actually messed up when typing the original post: range quantity is the same as target (25.6 m).

 

[pill]

Is your problem that you have been told the initial velocity is 16 and you don't know how it was calculated?

For the most part, yes. I used an online simulation for this lab and just continually changed the value of the initial velocity until it hit the target.

 

[PeterO]

For the most part, yes. I used an online simulation for this lab and just continually changed the value of the initial velocity until it hit the target.

I we look at your original data
Target is 25.6 m away from canon
Initial Velocity = 16 m/s
Time = 2.6 s
Range = 26.7m
No air resistance

and just ignire that initial velocity figure, we know that it covered 25.6 m ( horizintally) in 2.6 seconds - so the horizontal component will be a little less than 10m/s

SInce it took 2.6 seconds to get there, it spent 1.3 seconds gaining height, then 1.3 seconds coming back down.
Standard analysis of vertical motion will tell you how fast it was traveling vertically. Add those two components together using pythagorus and you will get the answer - who knows; it might even be 16 !

 

[pill]

I we look at your original data
Target is 25.6 m away from canon
Initial Velocity = 16 m/s
Time = 2.6 s
Range = 26.7m
No air resistance

and just ignire that initial velocity figure, we know that it covered 25.6 m ( horizintally) in 2.6 seconds - so the horizontal component will be a little less than 10m/s

SInce it took 2.6 seconds to get there, it spent 1.3 seconds gaining height, then 1.3 seconds coming back down.
Standard analysis of vertical motion will tell you how fast it was traveling vertically. Add those two components together using pythagorus and you will get the answer - who knows; it might even be 16 !

Thanks! This made is a bit more clear to me. I found got this equation, and your description made it make sense:
Vx=16cos(∡)=26.7m/2.6s
Vy=√(256-(26.7m/2.6s)²)
V=16=√(26.7m/2.6s)²+Vy²

 

[moxi]

Hi, I'm having the same issue and was wondering where you found your final formula? Any help is greatly appreciated!