How Do Projectile Motion Concepts Apply to Real-World Physics Problems?

  • Thread starter Thread starter pokemonstar
  • Start date Start date
Click For Summary

Homework Help Overview

The discussion revolves around the application of projectile motion concepts to real-world physics problems, specifically focusing on the time a projectile is in the air and the relationship between horizontal and vertical motion. Participants reference a PDF document and engage with equations related to motion under constant acceleration.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants inquire about calculating the time a ball is in the air and how to represent its position graphically over time. Questions arise regarding the relationship between horizontal and vertical motions and how to apply relevant equations to these scenarios.

Discussion Status

Some participants have offered guidance on using specific equations for motion, while others express confusion about the application of these equations and seek clarification on the concepts involved. Multiple interpretations of the problem are being explored, particularly regarding the effects of horizontal and vertical motion on the projectile's trajectory.

Contextual Notes

Participants mention a lack of thorough explanation of the equations provided by their teacher, which contributes to their uncertainty in applying the concepts to the problems at hand.

pokemonstar
Messages
2
Reaction score
0
http://www.whitwellhigh.com/jcantrell/cp/conceptdev/cpcd0501.pdf

I'm looking at only the 1st page of the pdf file above

#1 How do i figure out the # of seconds the ball is in the air?
I have to draw the positions...so how do i know how many meters it is after 1 second?

for #2 should the ball be touching the ground after 4 seconds? how do i know how steep to draw the curve?

how can i explain the question to #2 (motion affected by horizontal/vertical)

I'm probably looking too much into this. Someone please explain :x
 
Last edited by a moderator:
Physics news on Phys.org
Did you see the formula s = (1/2) a t2 before?

#2 will become much easier once you start with the last part of the question (how do the horizontal and vertical motion affect each other?). The answer, which you are going to remember for the rest of your (educational) life, is: they don't. The ball just keeps going horizontally at the same speed, no matter what the vertical motion is. So if you want to solve a physical problem, you can use two completely separate sets of equation (for uniform motion horizontally, and for uniformly accelerated motion due to gravity in the vertical direction).

Will that get you started?
 
CompuChip said:
Did you see the formula s = (1/2) a t2 before?

#2 will become much easier once you start with the last part of the question (how do the horizontal and vertical motion affect each other?). The answer, which you are going to remember for the rest of your (educational) life, is: they don't. The ball just keeps going horizontally at the same speed, no matter what the vertical motion is. So if you want to solve a physical problem, you can use two completely separate sets of equation (for uniform motion horizontally, and for uniformly accelerated motion due to gravity in the vertical direction).

Will that get you started?

first off, thank you for your response! =)

the formula my class got from my teacher was x(t) = 1/2 a [tex]\Delta[/tex]t^2 + Vo[tex]\Delta[/tex]t + Xo

so...i don't know. I'm a bit confused she didn't explain any of the equations she gave us thoroughly..

from what I'm getting from you. I plugged in x = 1/2 (10m/s^2) (1s) which gives me 5 m/s

since 1cm:5m I drew the ball falling down every 1 cm. I said the ball was in the air for 16 seconds, I don't know how to mathematically display that.

I got 1cn/s (1/16cm) but that's giving me 1/16th of a second.

-----------

For #2 I drew the ball curving down at 1cm/s since you said the speed is not affected whether going straight down or being thrown. is that what you meant?

Thanks so much
 
The basic equation for motion with constant acceleration using time is [tex]x\left(t\right)=x_0+v_0t+\frac{at^2}{2}[/tex].

x0= the initial position
v0= the initial velocity
a = the acceleration
t = the time

This gives us the position after a certain time t. x0 is where the object starts. (v0+at/2)t gives us the displacement due to the velocity.

For two dimensional motion (without air resistance), use the parametric equations:
[tex]x\left(t\right)=x_0+v_{0x}t[/tex]
[tex]y\left(t\right)=y_0+v_{0y}+\frac{at^2}{2}[/tex]
 
pokemonstar said:
http://www.whitwellhigh.com/jcantrell/cp/conceptdev/cpcd0501.pdf

I'm looking at only the 1st page of the pdf file above

#1 How do i figure out the # of seconds the ball is in the air?
I have to draw the positions...so how do i know how many meters it is after 1 second?

use d=0.5a(t)^2
the ball should fall to the ground from being dropped in the same time it would take the projectile to hit the ground. You know d (displacement) and a (acceleration due to gravity)

for how many metres after 1 sec sub in 1 for time


for #2 should the ball be touching the ground after 4 seconds? how do i know how steep to draw the curve?

how can i explain the question to #2 (motion affected by horizontal/vertical)

horizontal velocity is constant. Vertically there is acceleration. so one point to another going directly horizontally they should be the same distance. Going straigght down however the distance should be increasing.
I'm probably looking too much into this. Someone please explain :x

...
 
Last edited by a moderator:

Similar threads

Replies
19
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 22 ·
Replies
22
Views
5K
  • · Replies 21 ·
Replies
21
Views
2K
Replies
7
Views
1K
Replies
15
Views
3K