Asked by Nicholle
A skier takes off a ski jump, with a velocity of 25m/s at an angle of 30 degrees above the horizontal/ the take off point is 2 m above the slope, which falls away in front of her 15 degrees below the horizontal. Treating the skier as a point particle and ignoring air resistance,at what distance down the slope does she land?
Answers
Answered by
bobpursley
horizontal veloicty: 25cos30
vertical velocity: 25sin30
now, consider the slope. Making the ski starting point zero.
vertical distance to slope=horizantal distance(t)*sin15
vertical position of skis: 2+viv*t-4.9t^2
so when the veritical position of the ski is the same as the vertical position of the slope, it hits.
solution for time in air:
2+viv*t-4.9t^2=horizantal distance(t)*sin15
where horizonal distance is 25cos30*t
and viv is 25sin30
so put that all together and solve for time in air t.
Then the distance down the slope is solved from slope distance=horizontaldistance/cos30
vertical velocity: 25sin30
now, consider the slope. Making the ski starting point zero.
vertical distance to slope=horizantal distance(t)*sin15
vertical position of skis: 2+viv*t-4.9t^2
so when the veritical position of the ski is the same as the vertical position of the slope, it hits.
solution for time in air:
2+viv*t-4.9t^2=horizantal distance(t)*sin15
where horizonal distance is 25cos30*t
and viv is 25sin30
so put that all together and solve for time in air t.
Then the distance down the slope is solved from slope distance=horizontaldistance/cos30
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