A mountain climber jumps a 1.9-m-wide crevasse by leaping horizontally with a speed of 9.0 m/s ... If the climber's direction of motion on landing is - 45°, what is the height difference between the two sides of the crevasse?

Question

Physics

Marvin Nunez

29 October, 17:20

A mountain climber jumps a 1.9-m-wide crevasse by leaping horizontally with a speed of 9.0 m/s ... If the climber's direction of motion on landing is - 45°, what is the height difference between the two sides of the crevasse?

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Charlee Kaufman · Answer 1

The x component of velocity is 9.0m/s, y component is zero initially. a = - 9.8m / s2

Since the jumper lands at a - 45-degree angle the final Vy must have the same magnitude as the final Vx. V f2 = V 02 + 2ad Plugging in values: (-9.0m/s) 2 = 0m/s+2 (-9.8m / s2) d The height difference, d is equal to |-4.13| or 4.13 m.