# Velocity of bullet fired from a gun

attached below

## Relevant Equations:

final velocity=initial velocity + accelerationxtime
distance=((initial velocity+final velocity)/2)x time I first calculated the velocity v:
√2.8^2+6.3^2= 6.8942

then i used it as the final velocity, so final velocity=6.8942
and the initial velocity=0
acceleration=9.8

Then i substituted them into this equation:
final velocity=initial velocity + accelerationxtime
then time=0.703489843

hence i substituted all values into this equation:
distance= ((initial velocity+final velocity)/2) x time

so, distance=2.42499

but the answer is not correct...I wonder what's wrong T.T

Related Introductory Physics Homework Help News on Phys.org
cnh1995
Homework Helper
Gold Member
[QUOTE="jamiebean, post: 6344729, member:

and the initial velocity=0
[/QUOTE]
How?
A gun that fires a bullet with 0 initial velocity?

[QUOTE="jamiebean, post: 6344729, member:

and the initial velocity=0
How?
A gun that fires a bullet with 0 initial velocity?
[/QUOTE]
opps.. so initial velocity=6.8942?

jbriggs444
Homework Helper
2019 Award
opps.. so initial velocity=6.8942?
The initial velocity is a vector. It has a direction. It has components. Those components are given.

then i used it as the final velocity, so final velocity=6.8942
and the initial velocity=0
acceleration=9.8

Then i substituted them into this equation:
final velocity=initial velocity + accelerationxtime
then time=0.703489843
The total velocity (not 'final' velocity) seems reasonable, but they you go and compute time as the time it would take a rock dropped to accelerate to that velocity, which is not what's going on here.

Edit: I should read the problem more carefully. Bullet fired from the ground, but upward. I took y to be east and x to be north. Oops!

Last edited:
Delta2
Homework Helper
Gold Member
The bullet does composite movement: uniform motion in the x-axis , and constant deceleration motion in the y-axis. You are using some formulas (like for example distance=((initial velocity+final velocity)/2)xtime) that are valid only in the case of a body that does movement in a straight line.

But the bullet's trajectory is actually a curve (parabola) and not a straight line. However if you study its motion separately in the x-axis (or in the y-axis) then you can consider that it moves in a straight line with regards to these axis.
SO:
Use the same formulas but separately for the movement in the x-axis and separately for the movement in the y-axis. Like for example when you use the formula
final velocity=initial velocity+acceleration x time you have to use it for the final and initial velocity in the y-axis only and the acceleration in the y-axis only as well.

haruspex
Homework Helper
Gold Member
I first calculated the velocity v:
√2.8^2+6.3^2= 6.8942
That was a backward step.
More usually, trajectory problems give you the launch speed and angle, and the first thing you have to do is to find the horizontal and vertical components. Here they have helped you by stating the velocity in terms of those components.

Because there is no drag, the horizontal and vertical motions are independent. Write equations for the coordinates at time t after launch.

You'll need to determine the slope. If, for example, the barrel angle is 45º above the line of sight from a rangefinfer , then gravity is operating through about .7071 (i.e. the cosine of the angle) of the straight-line distance to the target. https://en.wikipedia.org/wiki/Rifleman's_rule
Delta2 said:
But the bullet's trajectory is actually a curve (parabola) and not a straight line.
It's a straight line only when you're shooting straight up or straight down.

jbriggs444
• 