Projectile Motion — How far from the gun does the bullet land?

Click For Summary
SUMMARY

The discussion focuses on calculating the distance a bullet travels horizontally after being fired from a gun, given its initial velocity components of 8.4 m/s (horizontal) and 9.2 m/s (vertical), with an acceleration due to gravity of 9.8 m/s². Participants clarify that the horizontal and vertical motions are independent, and the time of flight can be determined from the vertical motion. The relevant equations include the time of flight and the range formula, R = (v₀² sin(2θ))/g, which requires the initial velocity and launch angle for further calculations.

PREREQUISITES
  • Understanding of projectile motion principles
  • Familiarity with kinematic equations
  • Knowledge of vector components in physics
  • Basic trigonometry for calculating angles
NEXT STEPS
  • Learn how to derive time of flight for projectile motion
  • Study the derivation and application of the range formula R = (v₀² sin(2θ))/g
  • Explore the effects of air resistance on projectile motion
  • Investigate the relationship between launch angle and range in projectile motion
USEFUL FOR

Students studying physics, educators teaching projectile motion concepts, and anyone interested in understanding the dynamics of bullet trajectories.

Tinkylo
Messages
6
Reaction score
1
Homework Statement
The bullet fired from a gun on the ground has a velocity v. The x-component of the velocity is 8.4 ms-1 and the y-component of the velocity is 9.2 ms-1. x is the horizontal axis and y is the vertical axis. What is the distance in m between the gun and the point where the bullet hits the ground? Acceleration due to gravity is 9.8ms-2. Assuming there is no air resistance during the bullet's flight.
Relevant Equations
Not sure
I don't know how to link the x-component and y-component together.
 

Attachments

  • 螢幕截圖 2020-02-21 下午3.52.14.png
    螢幕截圖 2020-02-21 下午3.52.14.png
    12.4 KB · Views: 241
Physics news on Phys.org
Tinkylo said:
I don't know how to link the x-component and y-component together.
Distance, start velocity, final velocity, time, acceleration. For which does the same value apply to both coordinates?
 
Tinkylo said:
I don't know how to link the x-component and y-component together.
As there is no air resistance, the vertical and horizontal components of velocity are not linked in the problem.Consider the problem as independent horizontal and vertical motions.
 
Tinkylo said:
Relevant Equations:: Not sure

I don't know how to link the x-component and y-component together.
Don't link those, they are giving you the vertical and horizontal components of the initial velocity in order to facilitate the problem.
Hits:
Purely vertical movement: decelerated while moving up / stop / accelerated while falling down.
Purely horizontal movement: non-accelerated and lasting as much as the up-down vertical movement.
 
Lnewqban said:
Don't link those
It's not clear what @Tinkylo means by linking them. My interpretation is finding something that links the horizontal and vertical equations. See my hint in post #2.
 
  • Like
Likes   Reactions: hutchphd
Tinkylo said:
Homework Statement:: The bullet fired from a gun on the ground has a velocity v. The x-component of the velocity is 8.4 ms-1 and the y-component of the velocity is 9.2 ms-1. x is the horizontal axis and y is the vertical axis. What is the distance in m between the gun and the point where the bullet hits the ground? Acceleration due to gravity is 9.8ms-2. Assuming there is no air resistance during the bullet's flight.
Relevant Equations:: Not sure

I don't know how to link the x-component and y-component together.
You can work it out from first principles from the equations for x and y as a function of t (time). This involves finding the value of t when y = 0 and then using that value of t in the equation for x. Or you can use the relationship between range, velocity and launch angle that you may have been given: ##R = \frac{v_0^2 \sin{2\theta}}{g}##. If you use this relationship, you will have to find ##v_0## and the launch angle. Which method do you wish to use?

AM
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
862
  • · Replies 2 ·
Replies
2
Views
2K
Replies
40
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 18 ·
Replies
18
Views
2K
Replies
7
Views
2K
  • · Replies 6 ·
Replies
6
Views
7K
  • · Replies 13 ·
Replies
13
Views
3K
Replies
4
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K