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

Lorenzo Conner

Physics

5 August, 08:26

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?

+5

Answers (1)

Know the Answer?

Quinn Vaughan · 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.