Finding initial velocity and time Please

Click For Summary
SUMMARY

The discussion focuses on calculating the initial velocity and time required for a missile to hit a target 650 km away when launched at a 20-degree angle. Participants clarify the use of kinematic equations, specifically the equations for constant acceleration and constant velocity. The final calculations yield an initial velocity of approximately 3149.61 m/s and a time of flight of about 219.619 seconds. The importance of maintaining precision in calculations without premature rounding is emphasized to avoid errors.

PREREQUISITES
  • Understanding of kinematic equations for constant acceleration and constant velocity
  • Knowledge of projectile motion principles
  • Familiarity with trigonometric functions (sine and cosine)
  • Basic algebra for solving equations with multiple variables
NEXT STEPS
  • Study the derivation and application of kinematic equations in projectile motion
  • Learn how to analyze projectile motion using vector components
  • Explore the effects of air resistance on projectile trajectories
  • Practice solving real-world problems involving angles and distances in physics
USEFUL FOR

Students studying physics, educators teaching projectile motion, and engineers involved in trajectory calculations for projectiles or missiles.

  • #31


or with the formula u told me to use, would be:
t = (650 x 10^3) / 3157 = 206 secs

Remember x=v_x t. You entered the total speed, but instead you need to enter v_x=v cos(20). That's why you get a different answer.
 
Physics news on Phys.org
  • #32


so which is the correct way to do it?
and what would u advise me to put as my final answer?
 
  • #33


Picking either equations will do, however always pick the easiest to minimize the errors. Secondly for optimal accuracy don't calculate the cos/sin then round them and then use them in further calculation. Just keep them as cos/sin. Same for the other variables if you do that you get the following answers.

v=3149.61 m/s
t=219.619 s

You see that rounding constantly during the calculation introduces errors.
 
  • #34


thank you so much for all your time, help, and patience![/color]
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
854
  • · Replies 3 ·
Replies
3
Views
1K
Replies
8
Views
2K
  • · Replies 3 ·
Replies
3
Views
986
  • · Replies 17 ·
Replies
17
Views
2K
Replies
11
Views
1K
Replies
3
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 9 ·
Replies
9
Views
1K