Answer:
a. i. Cannot reach height A, ii. Vb*cos(θ) b. Va = Vb*cos(θ)
Step-by-step explanation:
a.
i. We can use the concept of conservation of mechanical energy to solve this problem. Since we know that the car will be at rest at a height of A, the car will have some kinetic energy due to its velocity and some potential energy due to its height. Once the car is at its highest height off the track, the car cannot reach a height of A again since it will still be moving forward from being released. When all the kinetic energy is transformed into potential, the car would have to be at rest.
ii. When the car reaches its highest point off the track, we know that its y component of velocity must be zero. Our velocity will be the unit circle components of our initial velocity Vb.
<Vbsin(θ>,Vbcos(θ)> -> Vbsin(θ) = 0,
Velocity will by Vb*cos(θ)
b. The gravitational potential energy must be the same at both times the car reaches height A. Therefore, the velocity of our initial push must equal the velocity at which the car reaches height A again. In part ii, this value was 0, but in this case, it is Va.
Va = Vb*cos(θ)