Answer
The ball hits the floor a horizontal distance of $~~4.8~m~~$ from point A.
Work Step by Step
We can find an expression for the gravitational potential energy at a height of $6.0~m$:
$U_1 = Mgh$
$U_1 = (M)(9.8~m/s^2)(6.0~m)$
We can find an expression for the gravitational potential energy at a height of $2.0~m$:
$U_2 = Mgh$
$U_2 = (M)(9.8~m/s^2)(2.0~m)$
We can use conservation of energy to find the kinetic energy $K_2$ at a height of $2.0~m$:
$K_2+U_2 = K_1+U_1$
$K_2 = 0+U_1-U_2$
$K_2 = (M)(9.8~m/s^2)(6.0~m)-(M)(9.8~m/s^2)(2.0~m)$
$K_2 = (M)(9.8~m/s^2)(4.0~m)$
$K_2 = (39.2~M)~J$
The rotational inertia of a solid ball is $I = \frac{2}{5}MR^2$
We can find an expression for the total kinetic energy:
$K = \frac{1}{2}Mv^2+\frac{1}{2}I\omega^2$
$K = \frac{1}{2}Mv^2+\frac{1}{2}(\frac{2}{5}MR^2)(\frac{v}{R})^2$
$K = \frac{1}{2}Mv^2+\frac{1}{5}Mv^2$
$K = \frac{7}{10}Mv^2$
We can find the ball's speed as it leaves the track at a height of $2.0~m$:
$K = \frac{7}{10}Mv^2 = (39.2~M)$
$v^2 = \frac{(10)(39.2)}{7}$
$v = \sqrt{\frac{392}{7}}$
$v = 7.48~m/s$
We can find the time it takes the ball to fall $2.0~m$:
$y = \frac{1}{2}at^2$
$t = \sqrt{\frac{2y}{a}}$
$t = \sqrt{\frac{(2)(2.0~m)}{9.8~m/s^2}}$
$t = 0.639~s$
We can find the horizontal distance the ball travels in this time:
$x = v_x~t$
$x = (7.48~m/s)(0.639~s)$
$x = 4.8~m$
The ball hits the floor a horizontal distance of $~~4.8~m~~$ from point A.
