Gayle runs at a speed of 3.50 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 19.0 m? Gayle's mass is 50.0 kg, the sled has a mass of 5.00 kg and her brother has a mass of 30.0 kg.

It seems she runs on flat surface and you are looking for final horizontal velocity.

Horizontal acceleration is zero, regardless of the height.

HORIZONTAL

Initial momentum = mv = 50 . 3.5 = 175

Final mass = 50 + 5 + 30 = 85

Final Velocity = V

85 V = 175 (Conservation of Momentum)

V = 175/85

VERTICAL COMPONENT (If you are interested)

Potential energy = mgh

Total potential energy = 55.g.19 + 30. g. (19 - 5)

= 1465 g

= final kinetic energy = 1/2 m v^2 = 1/2 85 v^2

1/2 85 v^2 = 1465 g

I have to disagree with the 1st post. That one assumes that there is no increase in speed from coming down the hill.

Best to use conservation of energy, in fact that was the subject you gave the question.
Gayle's initial energy = kinetic + potential
Gayle's initial energy = (1/2)*50 kg*(2.5 m/s)^2 + 50 kg*9.8 m/s^2*19 m
Sled's initial energy = potential = 5 kg*9.8 m/s^2*19 m
Brother's initial energy = potential = 30 kg*0.8 m/s^2*(19 - 5) m

Figure all those terms out. You'll get units of kg*m^2/s^2. 1 of those is equivalent to 1 Joule. That's the total energy at the start. Call it TotalEnergy

Since the snow's friction is zero and there's no energy lost anywhere else either, the kinetic energy of all 3 masses at the bottom of the hill has to be equal to TotalEnergy.

So TotalEnergy = finalKE = (1/2)*(50+30+5)kg*v^2
Solve for v^2 and take Sqrt.