Question

How do I know what equations to use for projectile motion problems? (Or a really helpful explanation on what I'm looking for to solve projectile motion problems because I just don't understand)

From these 3 equations, I don't understand what's being done to change them so they fit x and y components:
The underscores are v subscript final and initial, the delta is the triangle, I hope this makes sense and thank you in advance
v_f = v_i + a(delta)t
x_f = x_i + v_i(t) + 1/2at^2
v_f^2 = v_i^2 + 2a(delta)x

Answer 1

In proj motion, you can break the problem into two part: horizontal (x) motion and vertical (y) motion.
a) Horizontal Motion
It is a linear motion with no acceleration. Equation:
\(v_x=\text{constant} \) and
\(x=v_xt\)

Answer 2

Thank you for replying, but hmm, I still don't feel like I really understand. I"m going to attach a picture, it's a question with the answers already. If it's possible I'd like you to explain the process in what it is I'm supposed to be doing to solve these questions, I'd like to understand how to do this. Thank you again.

Answer 3

Forgot to attach the file

Answer 4

Okay, then i'll continue with b) and explain that pic:

Answer 5

Thank you very much!

Answer 6

b) Vertical Motion
Gravity affect the motion here, giving vertical acceleration of -g. Equation:
\[y=y_0+v_{0y}t-\frac12 gt^2 \\ v_y=v_0-gt\] and optionally:
\[v_y^2=v_{0y}^2-2gy\]

Answer 7

Let'smove to part A now:
Given initial velocity v0, theta and g. Asked for th (time to reach max height)

Answer 8

You have any idea what's the condition for a particle to reach it's max height when thrown?

Answer 9

Do you mean when the vertical velocity is 0, that's when the max height is reached?

Answer 10

Yep, correct!
Now, we know:
- vertical vel is zero (v=0)
- want to find time (t)
- initial velocity ( \(v_{0y}\) )
which equation in vertical motion should we use then?

Answer 11

I want to say this one

vy=v0−gt (sorry I'm not too sure how to do the subscripts)

Answer 12

but rearranging it to

Answer 13

t= vy-v0/g

Answer 14

Good! But you missed the minus sign.\[t=\frac{v_{0y}-v_y}{g}\] Plug in all what we know to this equation (v0, vy)

Answer 15

Ah, for g? So \[t = 0 - v _{y}/ -9.8\] I'm not too sure about v_{y}

Answer 16

Okay, vy is 0 at max height. v0y isn't zero, and for g=9.8 (sice we already put the minus of -9.8 from the beginning in the equations).
As for v0y (and v0x), its value is:
\[v_{0y}=v_0\sin {\theta} \\ v_{0x}=v_0\cos{\theta} \]

Answer 17

Sorry, so v\[v _{0y}\] is always equal to \[v _{0y}\sin\]theta
which is the same as \[v\]

Answer 18

Vy

Answer 19

It goes like this:
\[ t=\frac{v_{0y}-v_y}{g} \\ t=\frac{v_0\sin{\theta}-0}{g} \\ t=\frac{v_0\sin{\theta}}{g} \]
Let me remark important things here:
1) v0y = v0 sin(theta)
2) vy =0
3) actually, v ISN'T equal to 0 (only stops in vertical motion, but still moving in x direction)

Answer 20

Earlier, I said
"- vertical vel is zero (v=0)"
Sorry for this, it should be
"- vertical vel is zero (vy=0)"

Answer 21

Okay, I think I'm slowly getting this and vy is only equal to 0 because we want the velocity at max height in order to find the time it takes to get to max height?

Answer 22

Very good. You understand the point.

Answer 23

And that's okay, no need to apologize for your mistakes, I'm already grateful for your time as it is, so thank you. So for Part B, I'm not to sure why the equation in the answer is multiplied by 2. Is it because it's taking into account the time it takes to reach max height and then fall from max height?

Answer 24

Yes, the time needed to reach max height is equal to time needed to fall from max height.
To hit the ground, proj need to reach max height and then fall. Right?

Answer 25

Yes, so I think I understand the reason for the 2 now

Answer 26

So for part C, I'm not sure at all why sin(theta) is squared and \[v _{0y}\] as well

Answer 27

Okay, for c we're dealing with these variables:
- v0
- theta
- g
- and y (or H)
Which equation of vertical motion you'll choose?

Answer 28

and also vy=0 (they said "to reach max height")

Answer 29

I'm going to go with \[v^2_{y} = v^2_{0y} - 2gy\]

Answer 30

which would turn into

Answer 31

\[v^2_{0y}/2g\] and we get rid of v^2y because it's 0, right?

Answer 32

yep, y=v0y^2/2g
And you already know what v0y is.

Answer 33

yep so that would be \[v^2_{0y}\sin \theta^2\] i think

Answer 34

yes, you do prove part 3 then :D

Answer 35

v0y^2=v0^2sin^2(theta)

Answer 36

Thank you! I understand that a lot more. So with Part D, and I guess B as well, why don't we use cos since wouldn't that be for the horizontal components and R is horizontal?

Answer 37

We do use cos if we want to find the horizontal distance R
R=v0x*tH
R=v0*cos(theta)*2*tH
But to find tH (time needed to reach max height), we just need to see the vertical movement of the object.

Answer 38

So is it because we're using the vertical components to find tR for part B that we're sticking with sign?

Answer 39

sticking with sign? which sign do you mean?

Answer 40

Sorry, I meant sin

Answer 41

So for part D, I'm a bit lost again on how that equation is gotten to

Answer 42

yes, the time needed to reach max height only depend on vertical velocity V0y, hence sin. Nothing to do with horizontal velocity, V0x.

Answer 43

Part D isn't included in your figure above.

Answer 44

Oh, I'm so sorry, give me one second

Answer 45

Sorry, i've to off ffor a whhile

Answer 46

It's okay, thank you so much for all that you've helped me with so far, I am eternally grateful, and I'm so sorry for taking up so much of your time. Thank you again so much!