Answer
The sphere travels up along the incline a distance of $~~3.43~m$
Work Step by Step
In part (a), we found that the expression for the total kinetic energy of a sphere moving at a speed $v$ is $K = \frac{7}{10}Mv^2$
The gravitational potential energy at the highest point on the incline will be equal to the kinetic energy at the bottom of the incline.
We can find the distance $d$ that the sphere travels along the incline:
$U = K$
$Mgh = \frac{7}{10}Mv^2$
$g~d~sin~\theta= \frac{7}{10}v^2$
$d = \frac{7~v^2}{10~g~sin~\theta}$
$d = \frac{(7)(4.90~m/s)^2}{(10)(9.8~m/s^2)~sin~30.0^{\circ}}$
$d = 3.43~m$
The sphere travels up along the incline a distance of $~~3.43~m$.
