Projectile Motion of rifle bullet

AI Thread Summary
The discussion focuses on calculating the flight time and speed of a bullet shot horizontally at a target 44m away, hitting 3.0 cm below the aim point. The vertical displacement of 0.03m is used to derive the flight time using the equation for accelerated motion, resulting in a flight time of approximately 0.078 seconds. The horizontal speed is then calculated using the horizontal motion equation, yielding a speed of about 564.103 m/s. Participants clarify that the flight time is needed to find the velocity, not the other way around. The problem is resolved successfully, demonstrating the application of projectile motion principles.
aligass2004
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Homework Statement


A rifle is aimed horizontally at a target 44m away. The bullet hits the target 3.0 cm below the aim point. a.) What was the bullet's flight time? b.) What was the bullet's speed as it left the barrel?


Homework Equations


Xf = Xi +Vixt + 1/2(a)(t^2)
Vfx = Vix + at
Vfx^2 = Vix^2 + 2ax(t)


The Attempt at a Solution


I have tried figuring this out. I know that the velocity needs to be found, and I know that the velocity then needs to broken into components. I'm just having problems without an angle given.
 
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The vertical displacement is 0.03m. You can use that to get the flight time. The bullet is shot horizontally. hence the angle is 0.
 
Ok, but how do you find the velocity?
 
aligass2004 said:
Ok, but how do you find the velocity?

Did you find the flight time?
 
Don't I need the velocity to find the flight time?
 
aligass2004 said:
Don't I need the velocity to find the flight time?

No you don't. You need the flight time to find the velocity. :wink:

What is the vertical displacement? write an equation for vertical displacement in terms of time.
 
Ok, you said the vertical displacement is .03m. So is the equation delta x = Vx(delta time)?
 
aligass2004 said:
Ok, you said the vertical displacement is .03m. So is the equation delta x = Vx(delta time)?

no. that equation won't work. it's accelerating vertically... So you need an equation for accelerated motion.
 
I finally figured it out. I used Yf = Yi + Vyi(t) + 1/2ay(t^2). Since the angle is zero because it's shot horizontally, the y component of velocity is zero, eliminating the middle term. So I just plugged in and got the time to be .078s. Then I used Xf = Xi + Vx(t). I found Vx to be 564.103 m/s. Thank you so much!
 
  • #10
aligass2004 said:
I finally figured it out. I used Yf = Yi + Vyi(t) + 1/2ay(t^2). Since the angle is zero because it's shot horizontally, the y component of velocity is zero, eliminating the middle term. So I just plugged in and got the time to be .078s. Then I used Xf = Xi + Vx(t). I found Vx to be 564.103 m/s. Thank you so much!

No prob. Good job!
 
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