Help Solving air drag projectile problem

Click For Summary
SUMMARY

The discussion focuses on solving a projectile motion problem influenced by air drag, specifically using the equation involving constants and exponential decay. The user presents a series of mathematical transformations leading to a logarithmic equation, ultimately seeking assistance in determining the time of flight for a projectile launched from a height, factoring in air resistance. The conversation highlights the need for clarity on the original problem statement to provide effective guidance.

PREREQUISITES
  • Understanding of projectile motion equations
  • Familiarity with exponential functions and logarithms
  • Knowledge of air resistance modeling in physics
  • Basic calculus for solving differential equations
NEXT STEPS
  • Research the derivation of projectile motion equations with air drag
  • Learn about numerical methods for solving differential equations
  • Explore the use of Python libraries like SciPy for simulating projectile motion
  • Study the impact of varying launch angles on projectile trajectories
USEFUL FOR

Students and professionals in physics, engineers working on projectile dynamics, and anyone interested in advanced mathematical modeling of motion affected by air resistance.

Matt Jacques
Messages
81
Reaction score
0
Since my other thread is lacking attention, perhaps it is more suitable here:

I inserted some fixed constants and multiplied out

(48/.5)(1-^e-.5t)(sin 45) + (9.8/.25)(1-(.5t) - e^-.5t) = 0

(96 - 96e^-.5t)(sin 45) + 39.2(1-(.5t)-e^-.5t) = 0

67.88 - 67.88^e-.5t + 39.2 - 19.6t - 19.6e^-.5 = 0

107.2 - 87.48e^-.5t - 19.6t = 0

107.2 - 19.6t = 87.48^e-.5t

log(107.2 - 19.6t) = log(87.48^e-.5t)

log107.2 - log19.6t = -.5tLog(87.48)

2.030194 - log19.6t = -.5t(1.94198)

1.045463 - log19.6t = -.5t

-(1.045463 - log19.6t = -.5t)

-1.045463 + log19.6t = .5t

log19.6t = .5t + 1.045463

10^(.5t + 1.045463) = 19.6t

This is where I am stuck.
 
Physics news on Phys.org
Matt,

Could you post the original problem, or a link to your other thread? I don't see what you're trying to do with this.
 

Similar threads

Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 11 ·
Replies
11
Views
3K
Replies
4
Views
13K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 53 ·
2
Replies
53
Views
8K
  • · Replies 6 ·
Replies
6
Views
11K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 4 ·
Replies
4
Views
9K