Question

A 64 kg skier leaves the end of a ski-jump ramp with a velocity of 27 m/s directed 25° above the horizontal. Suppose that as a result of air drag the skier returns to the ground with a speed of 25 m/s, landing 17 m vertically below the end of the ramp. From the launch to the return to the ground, by how much is the mechanical energy of the skier–Earth system reduced because of air drag?

Answer 1

Answer:

$1.40\times 10^{4}Joules$

Explanation:

The mechanical energy  is conservated in the absence of non conservative forces. In this case there is an air drag, wich will acount for the lost of energy. In other words:

$\frac{1}{2}mv_{i} ^{2} +mgh_{i}-E_{lost}=\frac{1}{2}mv_{f} ^{2} +mgh_{f}$

Solving for the Energy lost:

$E_{lost}=\frac{1}{2}mv_{i} ^{2} +mgh_{i}-\frac{1}{2}mv_{f} ^{2} -mgh_{f}$

And so

$E_{lost}=\frac{1}{2}m(v_{i} ^{2}-v_{f} ^{2}) +mg(h_{i}-h_{f})$

substituting with data

$E_{lost}=\frac{1}{2}(64)((27) ^{2}-(25) ^{2}) +(64)(9.8)(17)=1.40\times 10^{4}Joules$

Hope my answer helps.

Have a nice day!