A rifle aimed horizontally at a target fired a bullet which lands 2 cm below the aim point. If the rifle was 25 m away find a) the flight time and b) the velocity with which the bullet left the rifle.
I would first have to convert the 2 cm into m but what would be the equations i would use?
to find the time would I use
t = sqrt((2*h)/g))
I got sqrt(2(.02m)/9.8)
which equals .0638 seconds
for the velocity would i use
v= v initial + at where v intial is 0
v= (9.8)(.0638s) which equlas .625 m/s
did i do this correctly?
Your first part is exactly correct!
For the second part: think what you have. The gun shoots the bullet. You know how long in flight it is. And you know the distance away from the target it is. We are now just talkin about horizontal distance and velocity. So...can you get it now?
WOuld i use v of the final vertical = v intial +at?
vvf= v intial +at
I am not sure if this is right though....
It's alot simpler...
If you know time of travel, and you know distance...how do you get velocity?? Remember, there is NO acceleration, or deceleration in the horizontal direction!
is it v= d/t ?
v= 25m/ .063s
v= 396.83 m/s
Yes! Good job.
a) The flight time of the bullet can be found using the equation t = sqrt(2h/g), where h is the vertical displacement and g is the acceleration due to gravity. In this case, the bullet lands 2 cm below the aim point, so the displacement h = -0.02 m. Plugging in the values, we get t = sqrt((2*(-0.02 m))/9.8 m/s^2), which simplifies to t ≈ 0.064 s.
b) To find the velocity with which the bullet left the rifle, we can use the equation v = d/t, where v is the velocity, d is the horizontal distance traveled, and t is the flight time. In this case, the horizontal distance is given as 25 m and the flight time is approximately 0.064 s. Substituting these values, we get v = 25 m / 0.064 s ≈ 390.625 m/s.
Therefore, the flight time of the bullet is approximately 0.064 seconds, and the velocity with which it left the rifle is approximately 390.625 m/s.