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Projectile Motion- Horizontal Range and Time of Flight

  • Thread starter leah3000
  • Start date
  • #1
43
0

Homework Statement



A rifle is used to fire two bullets. The first shot is at an angle of 60° above the horizontal and the second at an angle of 45° above the horizontal. The speed of EACH bullet as it leaves the rifle is 200ms^1. For EACH bullet calculate the horizontal range and the corresponding time of flight.

Homework Equations



Horizontal Range: v^2sin2ө/ g

Time of Flight= 2vsinө/g

The Attempt at a Solution



Horizontal Range for 1st shot= (200^2)sin2(60)/9.81

=3531m

Time of Flight= 2(200)sin(60)/9.81

=35.3 secs

Horizontal Range for 2nd shot= (200^2)sin2(45)/9.81

=4077m

Time of Flight= 2(200)sin(45)/9.81

= 28.8 secs

I'm not sure if I'm supposed to use the 200 as v or if I'm to calculate v somehow from the given information.:confused:
 
Last edited:

Answers and Replies

  • #2
123
1
You have used the equations correctly. I guess you are just not sure how the equations work? Maybe reading the proofs of these equations would help.
 
  • #3
43
0
ok...i just keep thinking that since it left the rifle at 200ms^1that that would be the initial velocity, u. I wasn't too sure if I could use it in the equation as v. I thought that i some how had to calculate it but then there would be more than one unknowns when using the equatios of motion:redface:
 
  • #4
43
0
ok...i just keep thinking that since it left the rifle at 200ms^1that that would be the initial velocity, u. I wasn't too sure if I could use it in the equation as v. I thought that i some how had to calculate it but then there would be more than one unknowns when using the equations of motion:redface:
 
  • #5
ideasrule
Homework Helper
2,266
0
v represents initial velocity, so it should be 200 m/s.
 
  • #6
43
0
ok thank you
 

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