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?

I will be happy to critique your work: the time of flight can be found by the distance it fell.

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?

Explain Bot Explain Bot answered
9 months ago

Yes, you are on the right track with your calculations.

To determine the time of flight, you correctly used the equation:

t = āˆš(2h/g)

t is the time of flight,
h is the vertical distance the bullet fell (2 cm = 0.02 m), and
g is the acceleration due to gravity (approximately 9.8 m/s^2).

So, substituting the given values into the equation:

t = āˆš(2 * 0.02 m / 9.8 m/s^2)
t ā‰ˆ 0.0638 s

Therefore, your calculation for the flight time is correct.

Now, to determine the initial velocity of the bullet, you can use the equation of motion:

Ī”y = vā‚€y * t + (1/2) * g * t^2

Since the bullet was fired horizontally, the initial vertical velocity (vā‚€y) is zero. The equation simplifies to:

Ī”y = (1/2) * g * t^2

Substituting the known values:

0.02 m = (1/2) * 9.8 m/s^2 * (0.0638 s)^2

Simplifying the equation gives:

vā‚€y ā‰ˆ 0.625 m/s

Therefore, your calculation for the initial velocity of the bullet is correct.

Well done! You correctly determined the time of flight and the initial velocity of the bullet.