Projectile Motion: Horizontal distance

Click For Summary
SUMMARY

The discussion focuses on calculating the horizontal distance a bomb travels after being released from a jet flying at 250 m/s at a 20-degree angle, from a height of 200 meters. Key equations identified include the horizontal component of velocity (vix = 234.923 m/s) and vertical component (viy = 85.505 m/s). The relevant kinematic equations for projectile motion are y = yi + viyt + (at²)/2 and R (range) = vxt. The discussion emphasizes the importance of recognizing the variables involved and applying the correct equations for constant acceleration due to gravity.

PREREQUISITES
  • Understanding of projectile motion principles
  • Familiarity with kinematic equations
  • Knowledge of trigonometric functions for resolving components
  • Basic concepts of constant acceleration
NEXT STEPS
  • Study the derivation of kinematic equations for projectile motion
  • Learn how to apply trigonometric functions in physics problems
  • Explore the effects of air resistance on projectile motion
  • Investigate advanced projectile motion scenarios involving variable acceleration
USEFUL FOR

Students studying physics, particularly those focusing on kinematics and projectile motion, as well as educators looking for clear examples of applying equations in real-world scenarios.

coffeeandativ
Messages
3
Reaction score
0

Homework Statement


A jet flying at 250m/s at an angle of 20 degrees above the horizontal releases a bomb 200m above the ground, how far does the bomb travel horizontally before hitting the ground?


Homework Equations


vix = vicosthetha
viy= visinthetha
not sure what else...

The Attempt at a Solution


vix= 234.923ms
viy= 85.505 ms

This is all I've managed to figure out so far, does anyone know more relevant equations that may help me out?
 
Physics news on Phys.org
okay so you have got the component velocities which is always a good start.

next step is to recognise the equations you will need.

try y=yi+viyt+(at2)/2

and R(range)=vxt

from these u must realize the variables you are given, what you want(range) and the variables you need to find


okay in reread your question and saw that you specifically asked for an equation.

you should know that the equations i gave above can be derived from the equation for constant acceleration( in this case constant gravity).

this equation is x=xi+vit+at2/2

where x is th final DISPLACEMENT, xi is initial displacement tis time and a is the CONSTANT acceleration. this formula does not work with variable acceleration
 
Last edited:

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
5
Views
4K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
4K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 4 ·
Replies
4
Views
9K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 9 ·
Replies
9
Views
3K