Answer
The initial speed is $~~2.33~m/s$
Work Step by Step
The rotational inertia of a solid ball is $I = \frac{2}{5}MR^2$
We can find an expression for the total kinetic energy:
$K = \frac{1}{2}Mv^2+\frac{1}{2}I\omega^2$
$K = \frac{1}{2}Mv^2+\frac{1}{2}(\frac{2}{5}MR^2)(\frac{v}{R})^2$
$K = \frac{1}{2}Mv^2+\frac{1}{5}Mv^2$
$K = \frac{7}{10}Mv^2$
The gravitational potential energy at the highest point on the ramp is equal to the ball's kinetic energy at the bottom of the ramp. We can use conservation of energy to find the initial speed:
$K_1+U_1 = K_2+U_2$
$K_1+0 = 0+U_2$
$\frac{7}{10}Mv^2 = Mgh$
$v^2 = \frac{10gh}{7}$
$v = \sqrt{\frac{10gh}{7}}$
$v = \sqrt{\frac{(10)(9.8~m/s^2)(1.50~m)~sin~15.0^{\circ}}{7}}$
$v = 2.33~m/s$
The initial speed is $~~2.33~m/s$.
