Answer
The speed of the ball when it reaches the top of the pinball machine is $1.53~m/s$
Work Step by Step
We can find an expression for the total kinetic energy of the solid sphere:
$KE = KE_{tr}+KE_{rot}$
$KE = \frac{1}{2}Mv^2+\frac{1}{2}I~\omega^2$
$KE = \frac{1}{2}Mv^2+\frac{1}{2}(\frac{2}{5}MR^2)~(\frac{v}{R})^2$
$KE = \frac{1}{2}Mv^2+\frac{1}{5}Mv^2$
$KE = \frac{7}{10}Mv^2$
We can use conservation of energy to find the speed of the ball at the top of the ramp:
$KE_f+U_f = KE_0+U_0$
$\frac{7}{10}Mv_f^2+Mgh = \frac{7}{10}Mv_0^2+0$
$\frac{7}{10}Mv_f^2 = \frac{7}{10}Mv_0^2-Mgh$
$v_f^2 = v_0^2-\frac{10gh}{7}$
$v_f = \sqrt{v_0^2-\frac{10gh}{7}}$
$v_f = \sqrt{(2.20~m/s)^2-\frac{(10)(9.80~m/s^2)(2.05~m)~sin~5.00^{\circ}}{7}}$
$v_f = 1.53~m/s$
The speed of the ball when it reaches the top of the pinball machine is $1.53~m/s$.
