Horizontal distance travelled at given time

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Homework Help Overview

The discussion revolves around projectile motion, specifically calculating the horizontal distance traveled by an object thrown at a certain velocity and angle. The original poster provides specific values for distance, peak height, peak time, and time of flight but expresses confusion regarding the calculation of horizontal distance at peak height.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the implications of neglecting air resistance and the constancy of horizontal velocity. There are hints about using trigonometric functions and the relationship between time of flight and horizontal distance. Some participants question the understanding of sine and cosine in this context.

Discussion Status

Several participants have offered hints and guidance regarding the relationship between horizontal distance and time of flight, while others explore different aspects of the problem, including the need for further understanding of projectile motion concepts. There is no explicit consensus on the best approach yet.

Contextual Notes

The original poster clarifies that this is not a homework problem but rather a project related to a basketball game, which influences the assumptions made about air resistance and the calculations involved.

TeamChuck
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I need to figure out how far an object has traveled at a given time.

For example purposes the object will be thrown with a velocity of 30 m/s and at an angle of 30 degrees.

I've figured out the distance/peak height/peak time/time of flight but I don't quite understand how to figure out the horizontal distance traveled at the peak height.

Distance: 79.5m
Peak Height: 6.99m
Peak Time: 1.53s
Time of Flight: 3.06s

I can't seem to find the formula with an example. Any help?
 
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TeamChuck said:
I need to figure out how far an object has traveled at a given time.

For example purposes the object will be thrown with a velocity of 30 m/s and at an angle of 30 degrees.

I've figured out the distance/peak height/peak time/time of flight but I don't quite understand how to figure out the horizontal distance traveled at the peak height.

Distance: 79.5m
Peak Height: 6.99m
Peak Time: 1.53s
Time of Flight: 3.06s

I can't seem to find the formula with an example. Any help?

Welcome to the PF.

Hint: if you neglect air resistance, the horizontal velocity is constant.

BTW, is this for schoolwork? If so, I can move it to the Homework Help forums for you.
 
TeamChuck said:
I need to figure out how far an object has traveled at a given time.

For example purposes the object will be thrown with a velocity of 30 m/s and at an angle of 30 degrees.

I've figured out the distance/peak height/peak time/time of flight but I don't quite understand how to figure out the horizontal distance traveled at the peak height.

Distance: 79.5m
Peak Height: 6.99m
Peak Time: 1.53s
Time of Flight: 3.06s

I can't seem to find the formula with an example. Any help?

You're going to slap your head when you hear this but since you have neglected air resistance, there are no forces in the horizontal direction. Hint: use sin (theta).
 
Welcome to PF!

Hi TeamChuck! Welcome to PF! :smile:

The horizontal speed is constant (because horizontal acceleration is zero), so just use the time of flight that you've found. :wink:
 
Not homework. I'm building a very basic basketball game that requires 3 sets of x,y coordinates: start/peak/end.

Neglecting air resistance the horizontal distance would be...half the total distance?

Head slap...more like head meet wall. If this is right. I haven't done any physics since high school, which was ten years ago.
 
TeamChuck said:
Not homework. I'm building a very basic basketball game that requires 3 sets of x,y coordinates: start/peak/end.

Neglecting air resistance the horizontal distance would be...half the total distance?

Head slap...more like head meet wall. If this is right. I haven't done any physics since high school, which was ten years ago.

Is this real basketball, or a video game? You may not be able to neglect air resistance, depending on what you are trying to do.
 
TeamChuck said:
Not homework. I'm building a very basic basketball game that requires 3 sets of x,y coordinates: start/peak/end.

Neglecting air resistance the horizontal distance would be...half the total distance?

Head slap...more like head meet wall. If this is right. I haven't done any physics since high school, which was ten years ago.

Do you get what to do with sin(theta)?
 
No, not real. Simple flash game. The user is given a random position on the court and then shoots at the net. The only thing the user controls is the angle and power(velocity). All shots go directly at the basket, its just a matter of whether you under or over throw the basket.

Heh, or whether you hopefully get it in.
 
  • #10
And no, I don't know what sin(theta) is.
 
  • #11
TeamChuck said:
And no, I don't know what sin(theta) is.

You wouldn't use sin(theta) anyway...
 
  • #12
berkeman said:
You wouldn't use sin(theta) anyway...

My bad. I was using vertical as theta = 0. cos(theta)
 
  • #13
This question is related to my topic but along a different line.

What is the equation used to calculate the angle/velocity if you are given the starting x,y position and the x,y coordinates of the peak of the curve?

Is there even a formula for that? Or would you need to know the time it took to reach the peak as well?
 
  • #14
TeamChuck said:
This question is related to my topic but along a different line.

What is the equation used to calculate the angle/velocity if you are given the starting x,y position and the x,y coordinates of the peak of the curve?

Is there even a formula for that? Or would you need to know the time it took to reach the peak as well?

OK, let's back up. You need to start with F=ma where a=v'=x". Also, v=x'. All of these (with the exception of m) are vectors. You will need to understand a little about vectors and basic calculus to solve these equations. If you do not understand any calculus or the concept of a vector, then that is where you need to begin. Try wikipedia on Newton's Laws to get a grasp on these concepts if they are not familiar.
 
  • #15
I have a general understanding of Calculus and Newtons three laws.

Maybe if I restate my question. I've been trying to understand the calculations on my own but I'm having difficulty. I have the x,y(pt.A) starting coordinates and I have the peak x,y(pt.B) coordinates. Because resistance is negligeble, the ending x,y(pt.C) would be twice the distance between pt.A and pt.B.

I would like to know the initial force and the time that it would take to complete.
 
  • #16
TeamChuck said:
I have a general understanding of Calculus and Newtons three laws.

Maybe if I restate my question. I've been trying to understand the calculations on my own but I'm having difficulty. I have the x,y(pt.A) starting coordinates and I have the peak x,y(pt.B) coordinates. Because resistance is negligeble, the ending x,y(pt.C) would be twice the distance between pt.A and pt.B.

I would like to know the initial force and the time that it would take to complete.

What have you calculated? Can you write the equation of motion in terms of the x,y positions, velocities and accelerations?
 
  • #17
Starting point would be 0,0(pt.A), peak of the curve would be 50,50(pt.B) and the end point would be 100,0(pt.C). As far as calculations go, I'm not sure. Because I don't have the force/angle I need to work back from the peak/range. I just don't know how to do that.
 
  • #18
TeamChuck said:
Starting point would be 0,0(pt.A), peak of the curve would be 50,50(pt.B) and the end point would be 100,0(pt.C). As far as calculations go, I'm not sure. Because I don't have the force/angle I need to work back from the peak/range. I just don't know how to do that.

First, take the angle piece out of it and shoot the ball straight up. What is the force acting on the ball and how fast is it going at apogee?
 
  • #19
Well, I wasn't sure if 'apogee' was a typo or not but as best I can figure you mean the farthest distance from the earth? If that's correct, then 50.
 
  • #20
TeamChuck said:
Well, I wasn't sure if 'apogee' was a typo or not but as best I can figure you mean the farthest distance from the earth? If that's correct, then 50.

What is the vertical speed there? and the force?
 

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