3.852272sin(45) = 2.72396
3.852272cos(45) = 2.72396
These numbers are vxi and vyi (initial velocities in the x and y direction)?
Would I use the original 3.852272m/s to solve for V final in the following formula: Vf^2 = Vi^2 + 2gy?
How do I separate this into vertical and horizontal components? Would it be something along the lines of 3.8sin(45)? (I'm probably completely wrong that was just a guess)
1.5sin(45) = 1.06 (more accurate, used fully accurate number when doing calculations on the calculator)
PE = mgh
PE = m*-9.8*-1.06
10.3m = 7/10mv^2 (moment of inertia)
M's cancel out...
10.3 = 7/10v^2
10.3/ (7/10)
14.84 = v^2
√14.84 = 3.85227m/s = Vf
Is this correct? Should it be...
haruspex:
For time, I could manipulate these two formulas: Vyf = vyi + gt and Y = Vyit + 1/2gt^2. My teacher says to use the first formula but you must first solve for Vyf.
However, I'm finding it difficult to find a formula for velocity where I don't need Vi of Vf. I guess my question would...
Haruspex: I am not trying to make people "guess" my variables. My teacher is very unclear and I have a very hard time understanding the subject matter when he explains it and I unfortunately had to miss some class time which didn't help either. According to the formula sheet, he has it noted...
Homework Statement
In this problem, I'm rolling a ball off a ramp. The ramp is supposed to simulate a roof. The hypotenuse of the ramp is 1.5 meters and the angle is 45 degrees. The distance from the edge of the ramp to the ground is -5.43 meters. That is the information I'm given. From there...