Recent content by mykosfemme

I gave it a lot of consideration, but I don't know what the difference in setting up the equation would be. I'm really not trying to be difficult, I'm just very weak with this type of problem for some reason.

Sorry for delayed reply! I'm still quite confused, but here's another attempt: Ok, so... horizontal: x=0, a = 0 m/s^2, vi = 20 m/s x = xo + vot + 1/2 at^2 = > 0 = x0 +20(4) + (1/2)(0)(4)^2 x0 = -80 vertical vi = 0 m/s, h = ?, a = -9.8 x = xo + vot + 1/2at^2 x = 80 + 0(4) + 1/2(-9.8)(4)^2 x=...

I'm sorry, by 10 I meant 20, as in 20 m/s. I got the value confused with another similar problem I was looking at. 1. Conditions: I know finding the horizontal component is Vcostheta and the vertical is Vsintheta. For this problem, the stone is thrown horizontally and the only vertical force...

Gneill - my apologies! I will stick to the format in the future. phinds & gneill: I am not entirely sure. I tried using v = sqrt 2gh => 10 = sqrt 19.6h 100 = 19.6h^2 5.12 = h^2 h= 2.26 And this is not correct. I got closer using x = xo +vo +1/2 at^2: 0 = xo + 10 + 1/2*-9.8*5^2 0= xo + 10...

Hello all! I am prepping for my finals and going over some work I had problems with before. This one I am still struggling with finding for some reason. Any help would be appreciated! So here is the question: A stone is thrown horizontally with an initial speed of 20 m/s from the edge of a...