In the scenario below, how long is the ski jumper airborne, and where does the ski jumper land on the incline?
A ski jumper travels down a slope and leaves the ski track, moving horizontally to the left with a speed of 25 m/s. The landing incline below the skier falls off with a slope of 33 degrees.
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While landing the the ski jumper makes a slope of of 33 degree with the horizotal.
The ski jumper has the initial horizontal velocity of 25m/s. The horizontal velocity componet of the ski jumper is not affected by the gravity at any time during the fall. But the vertical velocity of the jumper increases every second by the acceleration due to gravity.
Therefore at the time of fall the slope of the trajectory must be (dy/dt )(dx/dt) = Vertical velocity / horizotal velocity = tan 33 degree.
Therefore the vertical velocity v = Horizontal velocity * tan33 deg = 25tan 33 deg .
v = 25tan 33.
Therefore the vertical velocity is gained from the gravitional acceleration g in time t seconds .
Therefore t = (25tan33)/g = (25tan33)/9.81 = 1.65 seconds.
Therefore the ski jumper was airborn 1.65 seconds nearly.
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