Question

Ok, can someone show me ....

Answer 1

I found:

Horizontal component of the velocity:
\(\LARGE V_x=11\cos(44)\approx 7.9\)

Vertical component of the velocity:
\(\LARGE V_x=11\sin(44)\approx 7.6\)
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How do I find the maximum height that the bsll goes above the ground?

Answer 2

bsll is supposed to say ball.

Answer 3

What's the question ?

Answer 4

How do I find the maximum height that the bAll goes above the ground?

Answer 5

Do you know your equations of motion?

Answer 6

Which one?

Answer 7

I might as well just say, no I don't....

Answer 8

i think we need to know the height of those ppl

Answer 9

at least we need to know the height of the person throwing the ball

Answer 10

This is projectile motion correct, so it's better if you label your vector quantities more precisely, so we're dealing with the y - direction here\[\vec a = \frac{ v_{yf}-v_{yi} }{ t }\] where \[v_{yf}=0\] is there more information to this, all you gave is a diagram?

Answer 11

f is final, i is initial.

just clarifying....

Answer 12

I deleted the question I am so bump
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Julie throws a ball to her friend Sarah. The ball leaves Julie's hand a distance 1.5 meters above the ground with an initial speed of 11 m/s at an angle 44 degrees; with respect to the horizontal. Sarah catches the ball 1.5 meters above the ground.

Answer 13

Wait, a is acceleration, but for maximum height, why do we need it?
(btw, sorry again and again for deleted the question.)

Answer 14

The key thing to observe is that the vertical component of velocity will be 0 when the ball reaches the maximum height

Answer 15

\[v_y = u_y - gt\]

set that equal to 0 and find the time at which that happens

Answer 16

a = g in this case

Answer 17

what is u_y ?

Answer 18

\(a = -g\)
because we take upwards as increasing position

Answer 19

\(u_y\) = initial vertical component of velocity

Answer 20

oh. my teacher labelled it \(\LARGE y_{0_{^{_{\huge _x}}}}\)

Answer 21

You can label it what ever, long as it makes sense to you and your teacher

Answer 22

or, idk maybe I am confusing labels. I suck at this:)

Answer 23

thats ridankulous
normally we use \(v\) and \(u\) for velocity
\(x\) and \(y\) for positions

Answer 24

Yeah it is I that mixed something up (as always)

Answer 25

we never used u, always v_0 xD

Answer 26

Oh yes V_(0y), that!

Answer 27

v = u + at

is how i still remember my kinematics eqns from highschool

Answer 28

that's much easier to

Answer 29

it doesn't matter
but we all need to agree on same thing..

Answer 30

\(\LARGE V_y=V_{0y}-gt\)

\(\LARGE 0=V_{0y}+9.81t\)

and what did you say set equal to 0? like this, or am I wrong again?

Answer 31

looks good
and do you know why you're setting that equal to 0

Answer 32

no wait, you flipped the sign

Answer 33

\(\LARGE V_y=V_{0y}-gt\)

\(\LARGE 0=V_{0y}\color{red}{-}9.81t\)

and what did you say set equal to 0? like this, or am I wrong again?

Answer 34

(I just want to do this the way my teacher does it, because I got enough confusion already for 100 seasons in advance....)

Answer 35

oh, you said -g, so I thought... ok will take a note

Answer 36

\(\LARGE 0=V_{0y}+9.81t\)

Answer 37

And I am setting it to 0, because V_y is the derivative of D_y (where D is displacement - in ase of y, the height)

Answer 38

So we are already having the derivative of the function that we want to maximize.
Right?

Answer 39

oh, -9.81 ... my bad.

Answer 40

\(\LARGE 0=V_{0y}-9.81t\)

Answer 41

remember this :
\(g\) is acceleration due to gravity, it equals approximately \(9.81\) on surface on earth, it is always positive.

Answer 42

wait, so no minus; plus?

Answer 43

\(a\) is acceleration, in our case \(a = -g = -9.81\)

Answer 44

\(\Large 0=V_{0y}\color{red}{\bf-}9.81t\)

Answer 45

so g is pos, -g is neg.
at least I know that. I am a pes(t) - imist.

Answer 46

\(a\) is negative because \(g\) is pointing downward
but we are taking "upwards" as increasing position

Answer 47

Ok, and then I just find cthe maximum of our unknown displacement of y, based on the given velocity component (which is the derivative of the displacement)

Answer 48

right, one thing at a time
first find the time at which vertical component of velocity becomes 0

Answer 49

t=0?

Answer 50

at t=0 ... V_(0y) is 0.