You're missing the displacement because that is what the problem is asking for. The displacement is the maximum height that the bullet reaches. Use that equation, with accleration being gravity.
EDIT: OK!
Friction always points in the direction opposite the net force (excluding friction): this makes intuitive sense. However, this is not always true. For example, let's say that a car that was going up on an inclined plane suddenly breaks. In this case, the only force other than friction that acts...
Thank you for your patience! I now see the perpendicular component and can solve the problem. I'm teaching myself physics right now, and good help is hard to come by. Again, I really appreciate it.
I'm sorry for the belated reply. I couldn't access the site for some reason.
The first part of your logic makes perfect sense. However, I'm still not clear on the second part. What is the component of accleration perpendicular to the plane? If there is a net accleration perpendicular to the...
Assuming the only force acting on the car is gravity (I'm going to say that the car has constant velocity), then the only force from the road against the car is weight cos(theta).
I've attached an image of my work.
I thought that the normal force, in this case, is the force of the ground against the car. This is why I have the normal force vector pointing toward the car.
And yes, I also think the normal force is weight cos(theta). However, the book I have states that the normal force is...
I'm sorry, but I'm still confused. How does that explain why the two cars have different normal forces (the only difference being that turning car has centripetal accleration)?
If a car is going up or down an inclined plane, the normal force on the car is just mg cos (theta).
However, if a car is turning on an inclined curve, the normal force is mg/cos(theta). Why are the two normal forces not the same?
To clarify, here is an example problem:
A highway curve...