Hello, I need some help with this physics problem. This was the only one I couldn't solve, and the teacher wouldn't help me because she wanted me to figure it out for myself, just like everyone else tried. (Although no one did) Here is the question: Type in imgur dot com slash a slash eddOA
Thanks.
My work:
Since I knew the time it takes to fall is 4m/9.8 m/s^2, I needed to find the velocity of the object at the end of the ramp, then multiply it to the time to find the horizontal distance. So, by using the equations that I learned at school, which is mgsin(theta) minus umgcos(theta) = ma. I needed acceleration. I got 2.77m/s^2, so I plugged it in to the vf^2 = vo^2 + 2ad equation, and got around 5.9 m/s, since the distance was the root of the horizontal distance squared plus the vertical distance on the ramp squared. (root(4.5^2 + 4.5^2)). I got around 2.4 m, but it was wrong apparently. Can someone tell me what I did wrong and guide me? Thank you.
Ok, I am lost on what you are trying to do. You computed the distance on the ramp correctly. Then you find acceleration from the equation you used. Now you find vf at the end of the ramp with the vf^2 equation.
Then you apparently stopped. As I read the question, you need to find the distance from the end of the ramp it hits the ground. You can figure the vertical velocity at the end of the ramp, so figure the time to hit (fall) to the ground. Knwing that time, find horizontal distance traveled in that time