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Finding initial velocity for projectile motion at 0 degrees

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  • #1
I am doing a projectile motion problem with my own results, and the unknown's of initial velocity and time. It is at 0 degrees, which I thought would mean I could find the vertical down of the projectile (where at an angle you find the vertical up and down because it goes up first). I have tried multiple ways but I am not sure if what I am doing is right.

The foam bullet is launched from a height of 1.266metres and travels a length of 6.55metres at 0°.

∴Vertical Down

Acceleration due to gravity (a) = -9.8m/s
Final velocity (v) = 0
Initial velocity (u) = ?
Time taken (t) = ?
Displacement down (s) = 1.266

Therefore I decided to use the motion equation (v^2)=(u^2)+2as
And when calculated results in u equalling 4.98m/s

And therefore distance/speed=t
6.55/4.98=1.32seconds

However, my problem when I get to this stage is that other foam bullets of slightly different weight etc have completely different distances, therefore different times, but with this calculation apparently the same velocity.

So my problem is I do not know whether I am on the right track or not, or have completely messed it up. Thankyou in advanced for your help!
 

Answers and Replies

  • #2
PeterO
Homework Helper
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I am doing a projectile motion problem with my own results, and the unknown's of initial velocity and time. It is at 0 degrees, which I thought would mean I could find the vertical down of the projectile (where at an angle you find the vertical up and down because it goes up first). I have tried multiple ways but I am not sure if what I am doing is right.

The foam bullet is launched from a height of 1.266metres and travels a length of 6.55metres at 0°.

∴Vertical Down

Acceleration due to gravity (a) = -9.8m/s
Final velocity (v) = 0
Initial velocity (u) = ?
Time taken (t) = ?
Displacement down (s) = 1.266

Therefore I decided to use the motion equation (v^2)=(u^2)+2as
And when calculated results in u equalling 4.98m/s

And therefore distance/speed=t
6.55/4.98=1.32seconds

However, my problem when I get to this stage is that other foam bullets of slightly different weight etc have completely different distances, therefore different times, but with this calculation apparently the same velocity.

So my problem is I do not know whether I am on the right track or not, or have completely messed it up. Thankyou in advanced for your help!
You don't seem to have allowed/accounted for air resistance. For objects of the same shape and speed, air resistance has less effect on an object of greater mass.

Try just dropping the "bullets" and see which one reaches the ground first. That will give you some idea of the influence of air resistance.


btw. an object dropped from 1.2m will take much less than 1 second to reach the ground, in the absence of air resistance. A projectile fired at 0o will take the same time - in the absence of air resistance.
 
  • #3
Sorry I forgot to mention the bullets are fired from the same gun each time and the weight difference is about 0.2 grams maximum and the distances are ranging from 5.99m to 8.87m.

I am not exactly sure how to incorporate air resistance into the equation, but there could also be an interference factor as well couldn't there? It is a gun like a basic nerf gun that the bullets are firing from.

And are my calculation correct or is there another factor I am missing or should have, I might be able to calculate the elastic potential energy but I am not certain if and how it would help.

Thankyou for your speedy reply.
 
  • #4
1,065
10
Therefore I decided to use the motion equation (v^2)=(u^2)+2as
And when calculated results in u equalling 4.98m/s

And therefore distance/speed=t
6.55/4.98=1.32seconds
------------------------------------------------
You are calculating final velocity not initial velocity u.
If a body is dropped, in 1 sec. it will travel 4.9m
 
  • #5
okkay, so if I change it so I am calculating final velocity, won't it be the same?
 
  • #6
1,065
10
You can't get identical initial velocity even for the same bullet and gun with different weight load.
Energy produced by the gun remain constant. E=1/2 mv2
Reduced mass means higher velocity.
 
  • #7
Yes, that makes sense. But how could that be incorporated to find the velocity? Could that be used instead of projectiles?
 
  • #8
1,065
10
Sorry i do not know about guns and bullets. Never touch a gun before.
Using ballistic pendulum is another way of finding the speed of a bullet.
 
  • #9
I just mean incorporating the energy into the physics equation to get a more accurate result. It is only a toy gun so all I know is what results don't sound right. So in terms of equations, would that result in an accurate answer?
 
  • #10
1,065
10
If the toy gun is using a spring as a method of propulsion then you can find the potential energy stored when the spring is fully compressed(cocked).
 
  • #11
So if I find this I can find the velocity using E=1/2mv^2 instead of projectile. However it should be similar to the projectile? And was my projectile calculation correct for 0 degrees or should I have incorporated the horizontal distance?

Thankyou very much for your help.
 

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