Solving Physics Questions: Speed, Height & Time of Flight

In summary, the student is seeking help with their physics questions, specifically determining the speed at which air resistance becomes significant for a baseball and finding the height and time the ball will be airborne if launched at that speed. There is some confusion about the equations and terminology used, and the student is seeking clarification on their approach.
  • #1
tucky
30
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!
 
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  • #2
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.
 
  • #3
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!
 

FAQ: Solving Physics Questions: Speed, Height & Time of Flight

1. What is the formula for calculating speed in physics?

The formula for calculating speed is speed = distance/time. This means that the distance traveled by an object divided by the time it took to travel that distance will give you the speed of the object.

2. How do you calculate height in physics?

The formula for calculating height is h = v02sin2θ/2g, where v0 is the initial velocity, θ is the angle of launch, and g is the acceleration due to gravity.

3. What is the equation for time of flight in physics?

The equation for time of flight is t = 2v0sinθ/g, where v0 is the initial velocity and θ is the angle of launch.

4. How does the angle of launch affect the time of flight?

The angle of launch affects the time of flight because it determines the initial velocity and vertical velocity of the object. A higher angle of launch will result in a longer time of flight as the object will have a greater vertical velocity.

5. Can you calculate the speed, height, and time of flight without knowing all the variables?

No, all three variables (speed, height, and time of flight) are interdependent and are needed to accurately calculate each other. If one variable is missing, the others cannot be calculated.

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