Gayle runs at 5.84 m/s and dives on a sled, initially at rest onto the top of a
ID: 2261940 • Letter: G
Question
Gayle runs at 5.84 m/s and dives on a sled,
initially at rest onto the top of a frictionless
snow covered hill. After she and the sled
have descended a vertical distance of 7.2 m,
her brother who is initially at rest hops on
her back and together they continue down the
hill.
The acceleration of gravity is 9.8 m/s2 .
What is their speed at the bottom if the
total vertical drop is 16.4 m? Gayle
Explanation / Answer
CASE 1: when Gayle and sled have descended a vertical distance of 7.2 m applying conservation of energy; initial K.E. of Gayle + gravitational P.E. of gayle & sled = final K.E. of Gayle & sled => 0.5*m1*v^2 + (m1 +m2)*g*h = 0.5*(m1 +m2)*v1^2 => 0.5*37.5*(5.84)^2 + (37.5 +6.26)*9.8*7.2 = 0.5*(37.5 +6.26)*v1^2 => v1 = 13.05 m/s CASE 2: when the brother also jumps and they descended further (16.4 - 7.2 = 9.2 m) again applying conservation of energy; initial K.E. of (Gayle +sled) at case 1 + gravitational P.E. of (gayle +sled+ brother) at case 1= final K.E. of (Gayle+ sled+brother) at bottom => 0.5*(m1 +m2)*v1^2 + (m1 +m2+m3)*g*h = 0.5*(m1 +m2+m3)*v2^2 => 0.5*(37.5+6.26)*(13.05)^2 + (37.5 +6.26 +41.6)*9.8*9.2 = 0.5*(37.5 +6.26 +41.6)*v2^2 => v2 = 16.36 m/s
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