Gayle runs at a speed of 3.70 m/s and dives on a sled, initially at rest on the top of a frictionless snow-covered hill. After she has descended a vertical distance of 5.00 m, her brother, who is initially at rest, hops on her back and together they continue down the hill. What is their speed at the bottom of the hill if the total vertical drop is 15.0 m? Gayle's mass is 45.0 kg, the sled has a mass of 5.15 kg and her brother has a mass of 30.0 kg.

To solve this problem, we need to use the principle of conservation of energy. The total mechanical energy of the system (potential energy + kinetic energy) is conserved in the absence of non-conservative forces like friction.

Step 1: Calculate the initial potential energy of Gayle and the sled. Potential energy (PE) = mgh where m is the mass, g is the acceleration due to gravity (9.8 m/s^2), and h is the height. PE = (45 kg + 5.15 kg) * 9.8 m/s^2 * 5 m = 2457 J

Step 2: Calculate the kinetic energy of Gayle and the sled after descending 5 m. Since the total mechanical energy is conserved, the kinetic energy (KE) at this point is equal to the initial potential energy. KE = 1/2 * m * v^2 where m is the mass and v is the velocity. Solving for v, we get v = sqrt((2 * KE) / m) = sqrt((2 * 2457 J) / (45 kg + 5.15 kg)) = 9.8 m/s

Step 3: Calculate the final potential energy when Gayle's brother hops on. The total mass now is the mass of Gayle, the sled, and her brother. The remaining height is 15 m - 5 m = 10 m. PE = mgh = (45 kg + 5.15 kg + 30 kg) * 9.8 m/s^2 * 10 m = 7840 J

Step 4: Calculate the final speed at the bottom of the hill. Again, since energy is conserved, the final kinetic energy is equal to the final potential energy. Solving for v, we get v = sqrt((2 * KE) / m) = sqrt((2 * 7840 J) / (45 kg + 5.15 kg + 30 kg)) = 14 m/s

So, their speed at the bottom of the hill is 14 m/s.