A 417 kg pig is standing at the top of a muddy hill on a rainy day. The hill is 100m long with a vertical drop of 37.4 m. If the pig slips and begins to slide down the hill, assuming a negligible amount of friction what will be the pigs speed at the bottom of the hill?

Only the vertical drop H matters. The pig's mass and hill slope length don't matter.

g*H = V^2/2
where V is the velocity at the bottom.

Solve for V

V = sqrt(2 g H)

To find the pig's speed at the bottom of the hill, we can use the principle of conservation of energy. The potential energy that the pig has at the top of the hill will be converted into kinetic energy at the bottom of the hill.

First, let's find the potential energy of the pig at the top of the hill. The potential energy (PE) is given by the formula:

PE = m * g * h

where m is the mass of the pig (417 kg), g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the vertical drop of the hill (37.4 m).

PE = 417 kg * 9.8 m/s^2 * 37.4 m
PE = 156,224.4 J

Now, using the principle of conservation of energy, we can equate the potential energy at the top of the hill to the kinetic energy at the bottom of the hill:

PE = KE
156,224.4 J = (1/2) * m * v^2

where v is the velocity or speed of the pig at the bottom of the hill.

Simplifying the equation:

v^2 = (2 * PE) / m
v^2 = (2 * 156,224.4 J) / 417 kg
v^2 = 747.27 m^2/s^2

Finally, taking the square root of both sides will give us the speed of the pig at the bottom of the hill:

v = √(747.27 m^2/s^2)
v ≈ 27.34 m/s

Therefore, the pig's speed at the bottom of the hill, assuming no friction, will be approximately 27.34 m/s.