Recent content by Shadow37

Thanks for replying.

Hold on a sec. Let me get this straight, vertical velocity = squareroot (2 x 9.8 x .15) Yes?

Please, do not take me as being lazy. I have sat on this problem for hours now, and it still does not work for me. I have looked online, and went through several chapters in my physics book. I am pretty sure the answer is quite simple, but for some foolish reason, I cannot connect the dots.

The use of SOHCAHTOA is permitted because our triangle has a 90° angle from the height of the fountain to the top of the arc. The θ is 50°, and the third angle is 40°. From SOHCAHTOA, I used the TOA portion. Tangent of θ = opposite over adjacent. tanθ = .15m / adjacent. Since θ is 50°, you can...

The problem goes like this. A water fountain sprays water at a 50 degree angle from the horizontal. The top of the arc is .15m high. What is the horizontal speed of the water? Using SOHCAHTOA, I have managed to find the distance the water flies, and the angles of the triangle the 50 degree...