Wow! I totally didn't see you could sub Dy in for VySin0... So Dx = VoCos0(sqrt(g/2dy),
So Vo would equal Vo = (Dx/Cos)/(sqrt(g/2dy)). I just know I am wrong...
Okay so I derived Vo = sqrt((0.5(Dx/cos0/sinO))g)
Correct? And yeah I need to derive it exactly like he did... Somehow -.- .
Anyways.. I feel like I'm getting somewhere! Thank you for helping me
Time should equal t = 2((Vo)(Sinθ))/g
Position for Dx should be (VoCosθ)(2((Vo)(Sinθ))/g) And Position for Dy should be VoSinθ(2((Vo)(Sinθ))/g, but won't the Dy position be 0?
How would I make this an equation? I feel so stupid...
I got the equation Vo = (Dx/cosθ)(sqrt(g/2Dy)) from my teacher, but he says I need to derive it. At the minimum speed, the monkey and the bullet will meet at (Dx,0).. I'm not sure how to fit the angles into the equation though... How do Sinθ and Cosθ play into it? And thank you guys so much for...
I figured out the first part! Displacement of bullet at Dx is Vo x Sin 0 x t - 1/2gt^2 which equals Dy - 1/2gt^2 and the displacement of the monkey is Dy - 1/2gt^2... which is the same as the bullet so they meet! I don't understand how you determine the minimum speed though, but I get the...
Obviously the bullet wouldn't hit the monkey (if didn't drop) if it was pointed directly at it as gravity would pull it down, so it wouldn't be direct, but I'm not really sure how to show this with algebra..
Ahhhh, sorry I'm wrong... Hopefully I can be right about this:
The equation for the bullet's X position would be X = (Vx)(Vy/9.8m/s^2), the bullet's Y position would be y = ((Vyo+Vyf)/2)(Vy/9.8m/s^2), and the equation for the monkey's vertical position is y = (Vo+Vi)/2)(Vf/9.8m/s^2)I hope that...
Homework Statement [/B]
Hey guys, so here is my question:
A huntsman fires a gun pointing at a monkey hanging on the branch of a tree. The huntsman and the base of the tree are on a flat horizontal. At the moment the huntsman fires the gun, the monkey will release it's grip on the branch of...