Question

The peak of the trajectory occurs at time t1. This is the point where the ball reaches its maximum height ymax. At the peak the ball switches from moving up to moving down, even as it continues to travel horizontally at a constant rate.

Part C

What are the values of the velocity vector components v1,x and v1,y (both in m/s) as well as the acceleration vector components a1,x and a1,y (both in m/s2)? Here the subscript 1 means that these are all at time t1.

A) 0, 0, 0, 0
B) 0, 0, 0, -9.80
C) 15.0, 0, 0, 0
D) 15.0, 0, 0, -9.80
E) 0, 26.0, 0, 0
F) 0, 26.0, 0, -9.80
G) 15.0, 26.0, 0, 0
H) 15.0, 26.0, 0, -9.80

Answer 1

Hi! Did you try narrowing down your options at all?

Answer 2

i thought it was G, but its wrong

Answer 3

So we need \(v_{1,x}\), \(v_{1,y}\), \(a_{1,x}\), and \(a_{1,y}\), right?

Answer 4

yea

Answer 5

So, what is the acceleration (a.k.a. change in velocity) cause by? It's always a force, but what force is that, here?

Answer 6

gravity? I Have no idea

Answer 7

Yes you do have an idea! :P You said it and you're right! :P
And that's about constant, and always in the negative \(y\) direction (the direction gravity pulls). So, you know your \(a_{1,y}\). Right?

Answer 8

Now we can look at \(a_{1,x}\). Are there any forces in the \(x\) direction?

Answer 9

a1y is what? it cant be =9.8 because that is not an option

Answer 10

Right, the options for \(a_{1,y}\) are either \(0\) or \(-9.8\), right?

Answer 11

im confused

Answer 12

Alright! So we have this \(\small \it{projectile}\), I don't know what it is, which in the air.

What forces are acting on it? Just gravity, downward. So that is the negative \(y\) direction. Gravity causes an acceleration of \(-9.8\ [m/s^2]\) in the \(y\) direction. That's just a number you'll come to memorize. Have you learned that yet? At Earth's surface, the acceleration due to gravity will always be about \(-9.8\ [m/s^2]\)?

Answer 13

is A1x velocity?

Answer 14

In most physics projectile motion problems, you can always assume that the object will be accelerating downward at \(9.8\ [m/s^2]\).

Which is to say \(-9.8\ [m/s^2]\), since we count "down" as the negative direction.

\(a_{1,x}\) is an "\(a\)" variable, and physicists (including students) use "\(a\)" variables to indicate acceleration. So \(a_{1,x}\) is the acceleration in the \(x\) direction when you're at the time "\(t_1\)," as the problem states.

Answer 15

okay...

Answer 16

is that 0, since it is a turning point?

Answer 17

That's 0, but because there is no acceleration in the \(x\) direction (horizontal). When you throw a ball or keys to someone, it doesn't slow down in the \(x\) direction, getting to them.

If you're running at 5 mph, and you toss your keys in the air, they're going 5 mph in the \(x\) direction too. And they don't slow down, they'll stay moving with you at 5 mph!

Answer 18

ok....

Answer 19

There is no acceleration in the \(x\) direction. Just because there is no force in that direction (except for air resistance, which is like nothing).

Answer 20

hmmm

Answer 21

So \(a_{1,x}\) is \(0\).

Answer 22

ok

Answer 23

is it: 0, 0, 0, -9.81

Answer 24

Nope! Do you know why?

Answer 25

In terms of how the projectile is moving?

Answer 26

nope

Answer 27

Well, the accelerations are right.

But something's wrong with the velocity.

With a \(v_{1,x}\) of \(0\), you're saying there is no speed in the horizontal direction. Is that true? Maybe parts A or B mention a horizontal velocity?

Answer 28

Part C
What are the values of the velocity vector components v1,x and v1,y (both in m/s) as well as the acceleration vector components a1,x and a1,y (both in m/s2)? Here the subscript 1 means that these are all at time t1.

15, 26, 0, -9.81

Answer 29

Part B
What are the values of the intial velocity vector components v0,x and v0,y (both in m/s) as well as the acceleration vector components a0,x and a0,y (both in m/s2)? Here the subscript 0 means "at time t0."
15.0, 26.0, 0, -9.80

Answer 30

So, the projectile has an \(x\) velocity of \(15\ [m/s]\)?

Answer 31

If "15.0, 26.0, 0, -9.80" is part B's answer...

Answer 32

yes thats part b answer, but i cont figure out c

Answer 33

I gotcha.

So let's just go through them one at a time. Do you know what \(v_{1,x}\) means? And, if you do, do you know what its value is?

Answer 34

initial velocity horizontal

Answer 35

Ah! It has that same value, but that's not its meaning.

Answer 36

oh

Answer 37

So.. do you know what the meaning of \(v_{1,x}\) is? For this problem?

Answer 38

no

Answer 39

Not an idea? Give me a wild guess, even if it's wrong.

Answer 40

i tried

Answer 41

....?

Answer 42

Did you look at the attachment?

Answer 43

yes

Answer 44

Did you see what was circled?

Answer 45

The \(v_1\)?

Answer 46

yea

Answer 47

So \(v_{1,x}\) is the \(x\) part of the \(v_{1}\).

Answer 48

ok

Answer 49

same as x1? position

Answer 50

And \(v_{1,y}\) is the \(y\) part of \(v_y\).

Answer 51

\(x_1\) is the horizontal distance traveled. It's on the picture only to indicate that it is the \(x\) distance when \(t=t_1\). That subscript of "\(1\)" is to indicate that we want to know the \(x\), \(v\), or \(a\) at \(t=t_1\).

Answer 52

ok

Answer 53

how about 0, 26, 0, -9.81

Answer 54

no no, i mean 26, 0, 0, -9.81

Answer 55

So, at \(t=t_1\), the projectile is as high as it's going to get. It stops its ascent. It's still being pulled to Earth, causing a \(-9.8\ [m/s^2]\) \(y\) (vertical) acceleration. There is no \(x\) (horizontal) acceleration, and the \(x\) (horizontal) velocity is unchanged from when it was first launched at its initial velocity (\(v_0,x\)).

But that's not an option!

Answer 56

yea they are options

Answer 57

F) 0, 26.0, 0, -9.80

Answer 58

You said,, "no no, i mean 26, 0, 0, -9.81."

Answer 59

That's not an option.

Answer 60

Haha...

Answer 61

well im lost, I cant figure it out, its not:
A) 0, 0, 0, 0
C) 15.0, 0, 0, 0
E) 0, 26.0, 0, 0
G) 15.0, 26.0, 0, 0
H) 15.0, 26.0, 0, -9.80

Answer 62

it is either B, D, F?

Answer 63

So you say that \(v_{1,x}=0\) and \(v_{1,y}=26\)?

Yep, it's one of those, for sure.

Answer 64

One of those, referring to B, D, and F...

Answer 65

I give up

Answer 66

Well then, it'll be impossible to find the answer.

Answer 67

alright then

Answer 68

It's your choice... But if you ask questions, I'll answer them, as long as you don't ask for the answer...

Answer 69

well i already go -10% what could be worse than that

Answer 70

-100%?

Answer 71

Your up 90% from that.

Answer 72

For future reference:\[v_{1,x}\equiv\text{horizontal velocity at the top of the projectile's ascent}\\v_{1,y}\equiv\text{vertical velocity at the top of the projectile's ascent}\\a_{1,x}\equiv\text{horizontal acceleration at the top of the projectile's ascent}\\a_{1,y}\equiv\text{vertical acceleration at the top of the projectile's ascent}\]

For all projectile motion that ignores air resistance:
\[\begin{matrix}v_{1,x}& v_{1,y}& a_{1,x}&a_{1,y}\\v_{i,x}&0&0&g\end{matrix}\]

Answer 73

\[\begin{matrix}v_{1,x}& v_{1,y}& a_{1,x}&a_{1,y}\end{matrix}\]\[\begin{matrix}v_{i,x} &&0&&0&&g\end{matrix}\]
\(g\) on Earth's surface is about \(-9.8\ [m/s^2]\)
\(v_{i,x}\) is the initial horizontal velocity.