A solid ball rolls smoothly from rest starting at height H = 28.7 m until it leaves the horizontal section at the end of the track, at height h = 5.10 m. How far horizontally from point P does the ball hit the floor?

To solve this problem, we will use conservation of energy, including rotational energy, as well as kinematic equations for projectile motion.

First, when the ball is rolling down the track, its total mechanical energy (ME) is conserved, where mechanical energy is the sum of the gravitational potential energy (PE), translational kinetic energy (KE), and rotational kinetic energy (RE). Initially, at the top of the track, the ball is at rest, so it has no kinetic energy.

ME₀ = ME

PE₀ = PE + KE + RE

Gravitational energy is weight (mg) times height (h). Translational kinetic energy is half the mass (m) times the square of the linear speed (v). And rotational kinetic energy is half the moment of inertia (I) times the square of the angular speed (ω).

mgh₀ = mgh + ½ mv² + ½ Iω²

For a solid sphere, moment of inertia is two-fifths the mass (m) times the square of the radius (R), and for rolling without slipping, the angular speed is equal to the linear speed (v) divided by the radius (R).

mgh₀ = mgh + ½ mv² + ½ (⅖ mR²) (v/R)²

mgh₀ = mgh + ½ mv² + ⅕ mv²

We can use our conservation of energy equation to find the speed that the ball leaves the track.

mgh₀ = mgh + ⁷/₁₀ mv²

gh₀ = gh + ⁷/₁₀ v²

g (h₀ − h) = ⁷/₁₀ v²

v² = ¹⁰/₇ g (h₀ − h)

Plugging in values:

v² = ¹⁰/₇ (9.8 m/s²) (28.7 m − 5.10 m)

v² = 330.4 m²/s²

v = 18.18 m/s

After leaving the track, the ball is in free fall. We can model the projectile motion using kinematic equations, also known as SUVAT equations. In this case, we will use the equation:

s = ut + ½ at²

where

s is the displacement (distance and direction)
u is the initial velocity (speed and direction)
a is the acceleration
t is the time

In the vertical direction, the ball's initial velocity is 0 m/s. Taking down to be positive, the ball's displacement is 5.10 m, and its acceleration is 9.8 m/s². The time it takes the ball to hit the floor is:

s = ut + ½ at²

5.10 m = (0 m/s) t + ½ (9.8 m/s²) t²

5.10 m = (4.9 m/s²) t²

t² = 1.041 s²

t = 1.020 s

In the horizontal direction, the ball's initial velocity is 18.18 m/s, and its acceleration is 0 m/s². After 1.020 seconds, the ball has traveled a horizontal distance of:

s = ut + ½ at²

s = (18.18 m/s) (1.020 s) + ½ (0 m/s²) (1.020 s)²

s = 18.5 m

Rounding to three significant figures, the ball hits the floor 18.5 meters from point P.