A ball slides up a frictionless ramp. It is then rolled without slipping and with the same initial velocity up another frictionless ramp (with the same slope angle). In which case does it reach a greater height, and why

Rolling case achieves greater height than sliding case

Step-by-step explanation:

For sliding ball:

- When balls slides up the ramp the kinetic energy is converted to gravitational potential energy.

- We have frictionless ramp, hence no loss due to friction.So the entire kinetic energy is converted into potential energy.

- The ball slides it only has translational kinetic energy as follows:

                                   ΔK.E = ΔP.E

                                   0.5*m*v^2 = m*g*h

                                    h = 0.5v^2 / g

For rolling ball:

- Its the same as the previous case but only difference is that there are two forms of kinetic energy translational and rotational. Thus the energy balance is:

                                  ΔK.E = ΔP.E

                                  0.5*m*v^2 + 0.5*I*w^2 = m*g*h

- Where I: moment of inertia of spherical ball = 2/5 *m*r^2

             w: Angular speed = v / r

                               0.5*m*v^2 + 0.2*m*v^2 = m*g*h

                               0.7v^2 = g*h

                               h = 0.7v^2 / g

- From both results we see that 0.7v^2/g for rolling case is greater than 0.5v^2/g sliding case.

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation: