A particle, starting from point A in the drawing at a height h0 = 2.0 m, is projected down the curved runway.?

Question

A particle, starting from point A in the drawing at a height h0 = 2.0 m, is projected down the curved runway. Upon leaving the runway at point B, the particle is traveling straight upward and reaches a height hf = 4.5 m above the floor before falling back down. Ignoring friction and air resistance, find the speed of the particle at point A.

Answer 1

no picture given, but I will assume point A is at the top of a ramp.

The simplest way to solve this is using conservation of energy.
The difference in potential energy at the height of the apogee versus the starting point is equal to the kinetic energy of the particle at point a

.5*m*v^2=m*g*(hf-h0)
v=sqrt(2*g*(hf-h0))
=sqrt(2*9.81*2.5)
=7 m/s

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