butterflydreamer
  • butterflydreamer
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?....
Mathematics
schrodinger
  • schrodinger
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butterflydreamer
  • butterflydreamer
Oh part c question 12 of the attachment :)
Rhystic
  • Rhystic
there is no attachment....
butterflydreamer
  • butterflydreamer
:/. I'll try add it again Can you see it now?
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amistre64
  • 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
  • butterflydreamer
hmm.. i used the same basic method as with a ball/rock/frisbee...
amistre64
  • 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
  • amistre64
in a vacuum cleaner they might ...
amistre64
  • amistre64
the derivative of position will give us velocity, but im not sure what math you are able to use
butterflydreamer
  • 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
  • butterflydreamer
by the way, i wrote it wrong and my "theta" is meant to be an alpha LOL
amistre64
  • amistre64
the x direction has no acceleration acting on it. so it is constant.
butterflydreamer
  • butterflydreamer
sorry but i don't quite understand what you mean... So am i wrong in saying v = 4 ?
amistre64
  • 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
  • butterflydreamer
okay. Yes gravity is a downwards motion. I'm just confused on how else i would solve for velocity.
butterflydreamer
  • butterflydreamer
*initial velocity
amistre64
  • amistre64
well, velocity id distance per time frame.
amistre64
  • 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
  • amistre64
the horizontal velocity is therefore: \[v_x=\frac{6}{1.5}\frac{m}{s}\]
butterflydreamer
  • butterflydreamer
which equals to 4m/s ...
amistre64
  • amistre64
our velocity with respect to y, accelerates since it is affect by an up/down force ... the force of gravity
amistre64
  • amistre64
yes, in part b we prove: v cos(a) = 4
butterflydreamer
  • butterflydreamer
i've already answered part a and b I just needed help with part c.
amistre64
  • amistre64
what is 'exact form' ?
phi
  • phi
the velocity can be represented as a vector you want the magnitude of this vector
amistre64
  • amistre64
you found the initial velocities in part b, what angle has an adjacent leg of 4, and an opposite leg of 15?
phi
  • phi
I would do sqrt( v^2 cos^2 + v^2 sin^2) = sqrt(4^2 +15^2)
amistre64
  • amistre64
the length of the vector, represents the intial velocity in the direction of the toss.
butterflydreamer
  • butterflydreamer
tan (a) = 15/4 (a) = 75 degrees (nearest degree) ?
phi
  • phi
yes
butterflydreamer
  • butterflydreamer
alright, so we just substitute in (a) into Vcos (a) = 4 ? So V = 4/(cos 75) ?
butterflydreamer
  • butterflydreamer
so then velocity = 15.5 ?
phi
  • phi
By "exact form" I think they want the sqr form
butterflydreamer
  • butterflydreamer
ohhh alright. Thank you!!! Sorry if i frustrated you guys , projectiles are confusing lol.
phi
  • phi
btw, how do you show v sin a = 15 I don't see how
butterflydreamer
  • 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
  • phi
thanks!
butterflydreamer
  • butterflydreamer
no problem LOL

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