Recent content by SammyLP250

I'm sorry I haven't replied back. It turns out I wasn't the only one in the class that had trouble with this problem and the teacher realized that and worked it out in class the next day. After looking over it a couple times, I understand what to do now. This problem is definitely not as simple...

Alright then so h = [(v0sinθ0d)/(v0cosθ0)] - (g(d/v0cosθ)2)/2 which become h = tanθ0d - g(d/v0cosθ0)2/2

Ahhh That's right. Thank you again. These small mistakes are going to be the death of me. h = [(v0yd)/v0x] - [g(d/v0x)2/2]

I'm stuck right here. But I think I am trying to minimize v0x

I'll set T = d/v0x and substitute T into the second equation. h = [(v0yd)/v0x] - g (d/v0x)2 So now I'm left with two unknowns! :biggrin:

Gotcha. And I understand why g is negative. So they will become d = v0xT h = v0yT - gT2

Why yes i did :-p thank you. And I did not know that, I assumed the ball was at max height at 3m So I get v0x = v0cosθ v0y = vosinθ and for distance x = v0xt y = v0yt + .5(9.81)t2

Please help. I think I may be over thinking this problem because it looks very simple. Homework Statement Determine the minimum initial velocity, v0, and corresponding angle, θ0 needed to kick the ball just over the 3m high fence. x = 6m y = 3m Homework Equations X: vx=v0x x = x0+v0xt v0x...

Thanks for replying! Could you explain to me what you mean by check the difference? I'm a little slow...

Homework Statement A 1.57 kg brick measures 20.0 cm x 8.00 cm x 5.50 cm. Taking the zero of potential energy when the brick lies on its broadest face. (a) What is the potential energy when the brick is standing on end? Note: You can treat the brick as though all its mass is concentrated at its...