Gayle runs at a speed of 3.90 m/s and dives on a sled, initially at rest on the

Question

Gayle runs at a speed of 3.90 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 47.5 kg, the sled has a mass of 5.10 kg and her brother has a mass of 30.0 kg

*please give the initial equations you used, and steps to the final answer. The answer is NOT 9.91 m/s.

Explanation / Answer

Here,

u = 3.90 m/s

h = 5 m

h1 = 15 m

let the speed of the Gayle at the 5 m below is v1

using third equation of motion

v1^2 = u^2 + 2 * g * h

v1^2 = 3.90^2 + 2 * 9.8 * 5

v1 = 10.64 m/s

after the brother hop on , velocity is v2

Using conservation of momentum

(47.5 + 5.10)* 10.64 = (47.5 + 5.1 + 30) * v2

v2 = 6.78 m/s

Now , at the bottom after 10 m

Let the speed is v3

v3^2 = 6.78^2 + 2 * 9.8 * 10

v3 = 15.6 m/s

the speed of combined at bottom is 15.6 m/s

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