Recent content by Psychros

I actually found that I did my arithmetic wrong solving for time (subtracted instead of addition). Thanks for the help!

sorry, that's what i meant when i put √(V0cos(θ)2+V0sin(θ)2), that it's = V0. I know that it doesn't change, but I'm not sure how to find its' magnitude in this situation.

Since speed is the magnitude of velocity, by components of velocity I imagine you mean Vx and Vy, or √(Vcos(θ)2+Vsin(θ)2)?

Oh yes, i made a false assumption about the y component. Do you mean speed as in magnitude of velocity? Also, should (vy)2 = (v0y)2 - g(Δy) have a -2g(Δy)? I apologize for being slow, but I'm not sure how to use this either since I don't know the change in height from the maximum

X component will be the same through it's entirety. y=y0+v0y(t)+½at2 or Vy=V0y+gt Vy=1.3V0y-9.81t But Theoretically if we were starting at the maximum, V0y would = 0, Vy/-9.81=t? I'm getting stuck down either avenue, I do think I'm missing something intuitive here..

Hmmm I don't think we've learned that yet..

That is brilliant, thank you. However, Without knowing distance (x), time, or hard value for velocity (since it's expressed as a relation to the initial velocity) I'm not sure how to go about this.

Homework Statement A cannonball is fired with initial speed v0 at an angle 30° above the horizontal from a height of 38.0 m above the ground. The projectile strikes the ground with a speed of 1.3v0. Find v0. (Ignore any effects due to air resistance.) Homework Equations Time= 0=v0T-½gT2...

Homework Statement A woman throws a ball at a vertical wall d = 3.4 m away. The ball is h = 1.7 m above ground when it leaves the woman's hand with an initial velocity of 11 m/s at 45°. When the ball hits the wall, the horizontal component of its velocity is reversed; the vertical component...