the 2 kg ball is fired from point a with an Vo of 10 m/s up the smooth inclined plane. The angle is 37 degrees, it gives me the planes dimentions the ball rolls on, it is 1.5 m high and 2 m wide. It asks the distance in the x it lands. Also the velocity with which it strikes.
I know what the answer is supposed to be, 8.53m+2m=10.53m and velocity of 10 m/s- same as original.
The Attempt at a Solution
I found Vy=10*sin37= 6 m/s and Vx=10*cos37=8 m/s
We are in the energy section of the chapter so I used KE=PE to find the max hieight= 1.83m, dont know why, just did. Then I used the speed as a function of time kinematic equation, setting Vfy=0, found the time to reach max height= .6116 s. So now that I know the time to get to the top of the plane and the max height, how can I find the distance it will go?
I used the position as a function of time, using yo as 1.83' and vy=6 a=-9.81 and solved for the time from the max height to the ground and found .649 s. I am wondering if I should find the acceleration in the X using the speed as a function of time eq??? HELP ME, im going in circles...