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butterflydreamer
 one year ago
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
 one year ago
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?....

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butterflydreamer
 one year ago
Best ResponseYou've already chosen the best response.1Oh part c question 12 of the attachment :)

Rhystic
 one year ago
Best ResponseYou've already chosen the best response.0there is no attachment....

butterflydreamer
 one year ago
Best ResponseYou've already chosen the best response.1:/. I'll try add it again Can you see it now?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1how do we prove part a? since a frisbee has different dynamics then say a ball or or a rock or an anvil.

butterflydreamer
 one year ago
Best ResponseYou've already chosen the best response.1hmm.. i used the same basic method as with a ball/rock/frisbee...

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1a 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
 one year ago
Best ResponseYou've already chosen the best response.1in a vacuum cleaner they might ...

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1the derivative of position will give us velocity, but im not sure what math you are able to use

butterflydreamer
 one year ago
Best ResponseYou've already chosen the best response.1I 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
 one year ago
Best ResponseYou've already chosen the best response.1by the way, i wrote it wrong and my "theta" is meant to be an alpha LOL

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1the x direction has no acceleration acting on it. so it is constant.

butterflydreamer
 one year ago
Best ResponseYou've already chosen the best response.1sorry but i don't quite understand what you mean... So am i wrong in saying v = 4 ?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1you 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
 one year ago
Best ResponseYou've already chosen the best response.1okay. Yes gravity is a downwards motion. I'm just confused on how else i would solve for velocity.

butterflydreamer
 one year ago
Best ResponseYou've already chosen the best response.1*initial velocity

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1well, velocity id distance per time frame.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1the 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
 one year ago
Best ResponseYou've already chosen the best response.1the horizontal velocity is therefore: \[v_x=\frac{6}{1.5}\frac{m}{s}\]

butterflydreamer
 one year ago
Best ResponseYou've already chosen the best response.1which equals to 4m/s ...

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1our velocity with respect to y, accelerates since it is affect by an up/down force ... the force of gravity

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1yes, in part b we prove: v cos(a) = 4

butterflydreamer
 one year ago
Best ResponseYou've already chosen the best response.1i've already answered part a and b I just needed help with part c.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1what is 'exact form' ?

phi
 one year ago
Best ResponseYou've already chosen the best response.1the velocity can be represented as a vector <vx, vy> you want the magnitude of this vector

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1you found the initial velocities in part b, what angle has an adjacent leg of 4, and an opposite leg of 15?

phi
 one year ago
Best ResponseYou've already chosen the best response.1I would do sqrt( v^2 cos^2 + v^2 sin^2) = sqrt(4^2 +15^2)

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1the length of the vector, represents the intial velocity in the direction of the toss.

butterflydreamer
 one year ago
Best ResponseYou've already chosen the best response.1tan (a) = 15/4 (a) = 75 degrees (nearest degree) ?

butterflydreamer
 one year ago
Best ResponseYou've already chosen the best response.1alright, so we just substitute in (a) into Vcos (a) = 4 ? So V = 4/(cos 75) ?

butterflydreamer
 one year ago
Best ResponseYou've already chosen the best response.1so then velocity = 15.5 ?

phi
 one year ago
Best ResponseYou've already chosen the best response.1By "exact form" I think they want the sqr form

butterflydreamer
 one year ago
Best ResponseYou've already chosen the best response.1ohhh alright. Thank you!!! Sorry if i frustrated you guys , projectiles are confusing lol.

phi
 one year ago
Best ResponseYou've already chosen the best response.1btw, how do you show v sin a = 15 I don't see how

butterflydreamer
 one year ago
Best ResponseYou've already chosen the best response.1Well, 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

butterflydreamer
 one year ago
Best ResponseYou've already chosen the best response.1no problem LOL
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