Solving Physics Questions: Speed, Height & Time of Flight

Click For Summary
SUMMARY

This discussion focuses on solving physics questions related to the speed, height, and time of flight of a baseball. The key calculations involve determining the speed at which air resistance becomes significant, which is calculated to be 36 m/s (or 80 mph) using the equation mg = ½ * Cd * A * pair * v^2. The potential energy (PE) and kinetic energy (KE) relationship is used to find the maximum height of 66.122 meters. The participant seeks clarification on the time of flight equation, particularly regarding the effects of air resistance and the interpretation of "significant" air resistance.

PREREQUISITES
  • Understanding of basic physics concepts such as kinetic energy (KE) and potential energy (PE)
  • Familiarity with the drag equation, including coefficients like Cd and parameters like air density
  • Knowledge of algebra and trigonometry for solving equations
  • Basic grasp of calculus concepts, particularly derivatives and integration
NEXT STEPS
  • Study the drag equation in detail, focusing on how air resistance affects projectile motion
  • Learn about terminal velocity and its implications in physics problems
  • Explore the integration of differential equations to solve motion problems with external forces
  • Review conservation of energy principles in the context of forces acting on objects
USEFUL FOR

Students in algebra/trig-based physics courses, educators teaching physics concepts, and anyone interested in understanding the dynamics of projectile motion and air resistance.

tucky
Messages
30
Reaction score
0
Hi Everyone! Like always, thank you for helping me figure out my last physics question! I greatly appreciate the help. Here are my next questions:
Q: Determine the speed at which a baseball must move in order for air resistance to be significant?
A: mg = ½ * Cd * A * pair * v^2 pair = density of air (1.2kg/m^3)
m = mass .1488kg
g = 9.8m/s^2
A=(pi*r^2)=3.14*.115m^2=.04154 m^2
Cd = .5 (sphere)

v= square root of ((2*.1488kg*9.8m/s^2)/(.5* 3.14*.04154 m^2* *1.2kg/m^3 ))
v = 36 m/s or 80mph

Q: If the ball were launched vertically into the air at this speed, how high will it go?

A: After finding the speed in the previous question; I would use this equation to find height:
KE = PE
PE = mgy
KE = 1/2mv^2
(½)(.1488kg)(36m/s)^2 = (.1488kg)(9.8m/s^2)y
66.122m=y

Q: how long will it be air born?

x = x0 + v0 *t + 1/2at^2
66.122m = 0m + 36m/s * t + -9.8m/s^2t
v =

I am stuck on this part and I don’t know if the rest of it is correct? Please help!
 
Physics news on Phys.org
Actually, I think the first problem is ambiguous. What is meant by "significant". From your equation, you seem to interpret that as meaning the force due to the air resistance is the same as the force due to gravity. If the ball were falling, I could see that: the ball is now no longer accelerating, but going at a constant speed. However, for the ball going up, that makes no sense: gravity and air resistance are both pointing downward.

Do you have any reason for equating the kinetic energy when the ball is at "terminal speed" downward with potential energy? You do NOT have "conservation of energy" here because there is an outside force (air resistance) acting on the ball.

What you could do is use force= -mg- ½ * Cd * A * pair * v2 (notice that they are both negative. mdv/dt= force and you have v2on the right hand side so you will have to integrate
dv/(-mg - ½ * Cd * A * pair * v^2)= dt.
 
still confused

HallsoIvy,

Thank you so much for your quick response. I hate to bother you, however I am still confused over my last post. My class is an algebra/trig based physics class, therefore, if d stands for the derivative, the equations baffled me. Also, was I on the right track with the first equation? Sorry for my confusion with physics! Thanks again for all your help!
 

Similar threads

  • · Replies 14 ·
Replies
14
Views
2K
Replies
6
Views
3K
Replies
4
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 14 ·
Replies
14
Views
4K
Replies
4
Views
3K