Projectile motion question. Please help me :). I'm stuck on part c. Initial velocity means at t=0, v=? Right?
So if x = vt cos alpha, and vcos alpha= 4
Then x= 4t.. So wouldn't you just differentiate x=4t to find the velocity?
So initial velocity = 4 ? Or is this a trick question?....

- butterflydreamer

- schrodinger

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- butterflydreamer

Oh part c question 12 of the attachment :)

- Rhystic

there is no attachment....

- butterflydreamer

:/. I'll try add it again
Can you see it now?

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## More answers

- amistre64

how do we prove part a? since a frisbee has different dynamics then say a ball or or a rock or an anvil.

- butterflydreamer

hmm.. i used the same basic method as with a ball/rock/frisbee...

- amistre64

a flying frisbee does not fall like a rock. the acceleration is altered by its motion of flight.
ive thrown a frisbee, and a rock ... they dont have the same trajectories.

- amistre64

in a vacuum cleaner they might ...

- amistre64

the derivative of position will give us velocity, but im not sure what math you are able to use

- butterflydreamer

I guess that's true but the question doesn't really mention that?
Using the same general method, i still managed to prove part a:
|dw:1433953611028:dw|

- butterflydreamer

by the way, i wrote it wrong and my "theta" is meant to be an alpha LOL

- amistre64

the x direction has no acceleration acting on it. so it is constant.

- butterflydreamer

sorry but i don't quite understand what you mean...
So am i wrong in saying v = 4 ?

- amistre64

you are wrong for saying x' = something that varies with time. assuming a frictionless setup.
you see, gravity pulls downward, not sideways. there is not force pushing this thing sideways. it is moving forward under its own inertia which was imparted to it at the start.

- butterflydreamer

okay. Yes gravity is a downwards motion. I'm just confused on how else i would solve for velocity.

- butterflydreamer

*initial velocity

- amistre64

well, velocity id distance per time frame.

- amistre64

the velocity is in reference to the distance covered, divided by the time it took to get there.
if i travel 20 miles, in 1 hour ... my velocity is 20 mph

- amistre64

the horizontal velocity is therefore: \[v_x=\frac{6}{1.5}\frac{m}{s}\]

- butterflydreamer

which equals to 4m/s ...

- amistre64

our velocity with respect to y, accelerates since it is affect by an up/down force ... the force of gravity

- amistre64

yes, in part b we prove: v cos(a) = 4

- butterflydreamer

i've already answered part a and b
I just needed help with part c.

- amistre64

what is 'exact form' ?

- phi

the velocity can be represented as a vector
you want the magnitude of this vector

- amistre64

you found the initial velocities in part b, what angle has an adjacent leg of 4, and an opposite leg of 15?

- phi

I would do sqrt( v^2 cos^2 + v^2 sin^2) = sqrt(4^2 +15^2)

- amistre64

the length of the vector, represents the intial velocity in the direction of the toss.

- butterflydreamer

tan (a) = 15/4
(a) = 75 degrees (nearest degree) ?

- phi

yes

- butterflydreamer

alright, so we just substitute in (a) into Vcos (a) = 4 ?
So V = 4/(cos 75) ?

- butterflydreamer

so then velocity = 15.5 ?

- phi

By "exact form" I think they want the sqr form

- butterflydreamer

ohhh alright. Thank you!!!
Sorry if i frustrated you guys , projectiles are confusing lol.

- phi

btw, how do you show v sin a = 15
I don't see how

- butterflydreamer

Well, max height occurs when dy/dt = 0
so since y = -5t^2 + vsin(a) t
dy/dt = -10t + vsin(a)
Then sub indy/dt = 0 , t = 1.5
0 = -10 (1.5) + vsin(a)
vsin(a) = 15

- phi

thanks!

- butterflydreamer

no problem LOL

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