Projectile Motion: Calculating Flight Time and Initial Speed

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A rifle aimed horizontally at a target 50m away results in the bullet hitting 2.0cm below the aim point, prompting calculations for flight time and initial speed. The discussion reveals that the bullet's flight time was calculated to be approximately 0.064 seconds using the appropriate kinematic equations. Participants emphasize the importance of understanding the relationship between gravity, time, and position to solve the problem. The next step involves calculating the bullet's speed using the distance traveled and the determined flight time. Overall, the conversation highlights the challenges of applying physics equations to real-world scenarios.
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A rifle is aimed horizontally at a target 50m away. The bullet hits the target 2.0cm below the aim point.

A. What was the bullet's fight time?
B. What was the bullet's speed as it left the barrel?

Please HELP!
 
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jarrellboog04 said:
A rifle is aimed horizontally at a target 50m away. The bullet hits the target 2.0cm below the aim point.

A. What was the bullet's fight time?
B. What was the bullet's speed as it left the barrel?

Please HELP!

Welcome to PF.

How long does it take to drop 2 cm?
 
It does not give that info
 
It does not give me the time
 
I have some equations but i am not sure how to use them
 
jarrellboog04 said:
It does not give me the time

Well then figure it out.

You know gravity.
You know the distance.
So ...
 
None of those are the equations i have for the problem i have the answer from the back of the book but that does not help if i do not no how to answer it i tried many differnet ways but still can not find the right way
 
jarrellboog04 said:
None of those are the equations i have for the problem i have the answer from the back of the book but that does not help if i do not no how to answer it i tried many differnet ways but still can not find the right way

Which of those equations relates gravity, time and position? That is of course if you dropped it from rest, Vo = 0?
 
  • #10
i changed the 2cm to .02m to get the same units
 
  • #11
jarrellboog04 said:
i changed the 2cm to .02m to get the same units

OK. That you have to do.

So which equation did you try?
 
  • #12
I did conversion. I jus figured out the answer to part a i got .064s
i used the equarition
y^f=y^i + (v^y)i delta (t)-1/2g(delta (t)2
 
  • #13
jarrellboog04 said:
I did conversion. I jus figured out the answer to part a i got .064s
i used the equarition
y^f=y^i + (v^y)i delta (t)-1/2g(delta (t)2

Right equation. Good result.

Now it traveled 50 m in that time.
How fast you figure it was going to do that?
 
  • #14
But now i am trying do part b but cant
 
  • #15
jarrellboog04 said:
But now i am trying do part b but cant

Now it traveled 50 m in that time.
How fast you figure it was going to do that?
 
  • #16
I am going to use this equation and see if i can figure it out
x^f=x^i+(v^x)i delta (t)
 
  • #17
jarrellboog04 said:
I am going to use this equation and see if i can figure it out
x^f=x^i+(v^x)i delta (t)

OK. that works. Xi is 0 of course.
 
  • #18
i got Xi to be .02m
 
  • #19
i got xi to .o2cm
 
  • #20
jarrellboog04 said:
i got Xi to be .02m

No. When it was aimed it was aimed toward 0. It dropped .02.
 
  • #21
u still can not figure it out
 
  • #22
jarrellboog04 said:
u still can not figure it out

Which component are you talking about?

The time times x-Velocity is 50m.

x-Velocity is 50m/.064s

This is not rocket science. Oh, wait a minute, it is.
 
  • #23
I meant to say i still can not fingure it out
 
  • #24
thanks
 
  • #25
this is hard
 
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