A particle, starting from point A in the drawing (the height at A is 3.00 m), is projected down the curved runway. Upon leaving the runway at point B, the particle is traveling straight upward and reaches a height of 7.48 above the floor before falling back down. Ignoring friction and air resistance, find the speed of the particle at point A.

vA = 9.37 m/s

Explanation:

Ignoring friction and air resistance, total mechanical energy must be conserved at any point, as follows:

       

At the point A, the energy of the particle is in part gravitational potential energy, and in part kinetic energy.So, we can write the following expressions for the initial energies:

       

       

Since the total mechanical energy must be conserved at any point, we can choose as the final point, the one at which the ball, traveling straight upward, reaches to the maximum height before starting to fall down.At this point, just barely before the ball changes direction, the speed of the ball is zero, so the final kinetic energy is zero too:

       

The final gravitational potential energy, choosing as the zero reference level the ground, can be written as follows:

       

Rearranging in (1) , we can write:

       

Replacing by the values from (2), (3), (4) and (5) in (6), we get:

       

Simplifying and  rearranging, we finally can solve (7) for ViA, as follows:

       

Kinetic energy = 1/2mv^2. since the mass is constant we can form a ratio of it's kinetic energy at 60 divided by that at 30 to see ke(60)/ke(30) = +(60^2)/30^2). this may be said to 300% increase.

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