You foolishly attempt to jump across a crevasse whose far side is W = 2.31 m awa

Question

You foolishly attempt to jump across a crevasse whose far side is W = 2.31 m away horizontally, and H = 1.56 m lower vertically.

What minimum horizontal velocity must you have to make it to the other side? (Answer in m/s, assume g = 10 m/s2.)

You foolishly attempt to jump across a crevasse whose far side is W = 2.31 m away horizontally, and H = 1.56 m lower vertically.

Assuming you have the minimum velocity you just calculated in the previous part of the question, what is your landing angle below the horizontal in degrees? (Your landing angle is the angle your velocity vector makes with the horizontal.)

Explanation / Answer

y direction

y = y0 + v0y t + 1/2 a t^2

0 = 1.56 - 0.5*10*t^2

t=0.559 s

x direction

2.31 = v*0.559

v=4.13 m/s

b) vx = 4.13 m/s

vy = g t = 10*0.559 = 5.59 m/s

so angle = arctan(5.59/4.13)=53.54 degrees

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