Help Projectile Motion- Firing of a gun

Click For Summary
SUMMARY

A projectile is fired from a gun with initial horizontal and vertical velocity components of 30 m/s and 40 m/s, respectively. It takes approximately 4.08 seconds to reach the highest point of its trajectory. The magnitude of the projectile's velocity just before striking the ground is 50 m/s, calculated using the Pythagorean theorem. The horizontal distance traveled before impact is 244.8 meters, determined by multiplying the total time of flight by the horizontal velocity.

PREREQUISITES
  • Understanding of projectile motion principles
  • Knowledge of kinematic equations
  • Familiarity with vector components of velocity
  • Basic grasp of gravitational acceleration (9.8 m/s²)
NEXT STEPS
  • Study the derivation of kinematic equations for projectile motion
  • Learn about the effects of air resistance on projectile trajectories
  • Explore advanced projectile motion simulations using software like MATLAB
  • Investigate real-world applications of projectile motion in sports and engineering
USEFUL FOR

Students studying physics, educators teaching mechanics, and anyone interested in understanding the principles of projectile motion and its applications.

fewill
Messages
4
Reaction score
0
1. A projectile is fired from a gun (on the ground)and has initial horizontal and vertical components of velocity equal to 30 m/s and 40 m/s, respectively.a) How long does it take the projectile to reach the highest point on the trajectory? b) What is the magnitude of the projectile's velocity just before striking the ground? c) How far does the projectile travel horizontally before it hits the ground?

Homework Equations





The Attempt at a Solution

a) t= vy-viy/g 0-40/9.8= 4.08s
b) v= square root of viy^2 + vix^2= 50
v=viy+gt 0-9.8*4.08=40 m/s
c) total time = 2*4.08, 30*8.16= 244.8 m
 
Physics news on Phys.org
fewill said:
1. A projectile is fired from a gun (on the ground)and has initial horizontal and vertical components of velocity equal to 30 m/s and 40 m/s, respectively.a) How long does it take the projectile to reach the highest point on the trajectory? b) What is the magnitude of the projectile's velocity just before striking the ground? c) How far does the projectile travel horizontally before it hits the ground?

Homework Equations





The Attempt at a Solution

a) t= vy-viy/g 0-40/9.8= 4.08s
b) v= square root of viy^2 + vix^2= 50
v=viy+gt 0-9.8*4.08=40 m/s
c) total time = 2*4.08, 30*8.16= 244.8 m

You and reverse should get together!

(a) Yes, at maximum height the y component of velocity is 0 and, at acceleration g, it will take (vy- viy)/g seconds to reach that height. (I don't know why you switched to 0-40 instead of 40-0 obviously the time is NOT negative.)

(b) You've actually calculated the speed (magnitude of velocity) initially but, yes, because of the symmetry that is the same as the speed at the end of the trajectory.

(c) I don't know why you wrote "v=viy+gt 0-9.8*4.08=40 m/s". Obviously, 0-9.8*4.08= -40 m/s, not 40 m/s, but in any case, that is the vertical component of speed just as the projectile hits the ground- which you were not asked. Yes, again because of the symmetry of the trajectory, the total time is exactly twice the time to maximum height and the horizontal distance is just the horizontal component of velocity multiplied by that time.
 

Similar threads

Replies
40
Views
2K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 24 ·
Replies
24
Views
3K
Replies
18
Views
2K
Replies
11
Views
1K
Replies
15
Views
3K