glenohumeral13 said:
Could I find that with the given information?
I'm not sure... Let's see... The final velocity of the ball will be the same in free fall or rolling with no resistance along the incline. That velocity would be v = SQR (2 * 9,81 * 6 m * sin 24º) = 6,91 m/s. You state that there is an extra initial velocity of 1 m/s, so we would have a total of 7,91 m/s.
But 5 m/s is mentioned in the problem as final velocity. Thus, there is a braking force. It could come from rolling resistance and from rotational kinetic energy acquired by the ball during its run... Rolling resistance is a function of the mass of the ball, g, the angle and a coefficient mu. Rotational kinetic energy is a function of the angular velocity w, itself a funcion of v and the ball's radius r, and of the moment of inertia of a sphere 2/5 * m* r^2
Mass is not given. It might cancel away, I'm not sure, but the problem could perhaps be solved 'by energies', deriving the solution also in terms of the unknown magnitudes, maybe m (if it doesn't cancel away), a coefficient of rolling resistance mu, and the ball's radius r...
That, in case it can be solved at all... I am myself a solver of easy problems only...