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?

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. Vf2 = V02 + 2adPlugging in values:(−9.0m/s)2=0m/s+2(−9.8m/s2)dThe height difference,d is equal to |-4.13| or 4.13 m.

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aristocles aristocles · Answer 2 · 2019-10-17T02:57:05+02:00

Answer:

y = 4.13 m

Explanation:

Since the climber jumps horizontally with speed 9 m/s

so the time taken by the climber to cross the distance is given as

[tex]t = \frac{d_x}{v_x}[/tex]

[tex]t = \frac{1.9}{9}[/tex]

[tex]t = 0.211 s[/tex]

now we know that he land with a velocity at an angle of 45 degree

so we will have

[tex]v_y = v_x[/tex]

[tex]v_y = 9 m/s[/tex]

so we can find the vertical displacement by using kinematics

[tex]v_f^2 - v_i^2 = 2 a y[/tex]

[tex]9^2 - 0 = 2(9.81) y[/tex]

[tex]y = 4.13 m[/tex]