Question

i need someone to explain projectile motion to me :(

Answer 1

What do you want to know about it

Answer 2

everything...my god.

Answer 3

like vertical velocity ..calculate total time...range..height and final velocity

Answer 4

vertical velocity is the velocity you find in the vertical direction

Answer 5

and what do you mean by range?

Answer 6

Also you can mostly arrive to certain answers using the equations for projectile motion

Answer 7

yyea..i know the defs but i need to know the material

Answer 8

If you give me a few minutes, I will write a detailed lesson for you. Alright?

Answer 9

yes....that would be amazing

Answer 10

If motion is in a straight line, then the net force exerted on objects is along the line of motion. When an object is projected at an angle and the force of gravity is acting on the object, then the object's path is parabolic. In these situations, we must be able to differentiate between the horizontal and the vertical components of motion.

Vertical components:\[v _{y}=\pm v _{i}\sin \theta \]
\[\Delta d _{y}=v _{1_{y}}\Delta t \pm \frac{ 1 }{ 2 }g(\Delta t)^{2}\]whre g = 9.8 m/s², of course!

Horizontal Components:\[\Delta d _{x}=v _{x}\Delta t\]
\[v _{x}=v _{1}\cos \theta\]

Please keep in mind that for simple projectile motion, acceleration is only considered in the y direction due to gravity. Motion in the x direction is constant.

Range:
The distance a projectile travels in the horizontal (x) direction. To calculate the range simply use the distance equation in the x direction, given above.

Answer 11

@rayford you read the lesson I gave you while I think of and write an example for you. Cool?

Answer 12

yes :)

Answer 13

i'll have 2 look at it tomorrow..because i have to go 2 bed man.

Answer 14

No problem, I was almost done and my server reset so now I have to start all over again anyway. But do you understand the lesson at least? @rayford?

Answer 15

Example:
Suppose you throw a baseball at an angle of 60º to the horizontal, with an initial velocity of 84 m/s. How far will it travel?

Solution:
Given:\[v _{1}=84 m/s\]
\[\theta =60º \]
Required:\[\Delta d _{x}\]
\[\Delta t\]
Analysis:
Let's choose up and to the right as positive directions. Hence in the x direction\[v _{1_{x}}=v _{1}\cos \theta =(84 m/s)\cos 60º=42 m/s\]
\[\Delta d _{x}=v _{x}\Delta t\]
Similarly, in the y direction\[v _{1_{y}}=v _{1}\sin \theta =(84 m/s)\sin 60º=72.75 m/s\]
\[a _{y}=g =-9.8 m/s ^{2}\]because up was designated as positive, acceleration due to gravity is negative.

Since the baseball returns to the ground, its net distance is zero. Hence\[\Delta d _{y}=0\]
\[\Delta d _{y}=v _{1}\Delta t +\frac{ 1 }{ 2 }g(\Delta t)^{2}\]Solution:
We will first solve for time in the y direction. Hence\[0=(72.75 m/s)\Delta t - (4.9 m/s ^{2})(\Delta t)^{2}\]We can factor the time. Hence\[0=\Delta t(72.75-4.9\Delta t)\]\[\Delta t = 0\]or\[\Delta t =14.85\]Thus the range is\[\Delta d _{x}=v _{x}\Delta t =(42 m/s)(14.85 s)=623.7 m\]Statement:
The baseball travels 6.2 x 10² m.

Answer 16

i just got back to it! and yes i do understand it. thank you so much :)