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Part 1

Skiers launch from a ramp at a height of hh meters above the landing area, with the ramp angled at θ\theta degrees above the horizontal. Due to different skiing techniques and conditions, skiers will have launch speeds ranging from vminv_{\text{min}} to vmax m/sv_{\text{max}} \text{ m/s}. Your task is to determine the minimum width of the landing ramp needed to safely catch all skiers.

The landing ramp will have a very simple design. It will be a flat platform exactly on level at the ground.

Find the horizontal and vertical speed of the slowest potential skier. His speed magnitude should be vmin=26 m/sv_{\text{min}} = 26 \text{ m/s} and launch ramp incline is of θ=7°\theta = 7°.

Correct!

Solution:

To find the horizontal and vertical components of the slowest skier's velocity, we decompose the launch velocity vector using trigonometry.

The skier launches with speed vmin=26 m/sv_{\text{min}} = 26 \text{ m/s} at angle θ=7°\theta = 7° above the horizontal.

The horizontal component is: vmin,x=vmincos(θ)=26cos(7°)=25.81 m/sv_{\text{min}, x} = v_{\text{min}} \cos(\theta) = 26 \cos(7°) = 25.81 \text{ m/s}

The vertical component is: vmin,y=vminsin(θ)=26sin(7°)=3.17 m/sv_{\text{min}, y} = v_{\text{min}} \sin(\theta) = 26 \sin(7°) = 3.17 \text{ m/s}

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