Projectile Motion- Horizontal Range and Time of Flight

1. Mar 10, 2010

leah3000

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

Horizontal Range: v^2sin2ө/ g

Time of Flight= 2vsinө/g

3. 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.

Last edited: Mar 10, 2010
2. Mar 10, 2010

benhou

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. Mar 10, 2010

leah3000

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

4. Mar 10, 2010

leah3000

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

5. Mar 10, 2010

ideasrule

v represents initial velocity, so it should be 200 m/s.

6. Mar 10, 2010

ok thank you