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## Homework Statement

Long jumper jumps 8.12 m, reaching a height of 0.84 m half way through his jump. What was his velocity as he left the ground? Also, how far would he be able to jump on the moon (g = 1.63 m/s^2) and how much time would be spend off the lunar surface.

## Homework Equations

d = v1 + at^2 / 2

v = d/t

## The Attempt at a Solution

Ok, so I got the answer for the first part, but I just need a recheck from someone, The initial velocity as the jumper left the ground is 10 m/s in the horizontal direction and 0 in the vertical direction? The reason that I'm not sure of this answer is because when I find the angle (22 degrees) the answer becomes 9.27 m/s.

Now the second part is what I'm having problems in because the only variables known are the following:

Horizontal Direction:

V1 = 10 m/s

a = 0 m/s^2

Vertical Direction:

V1 = 0 m/s

a = 1.63 m/s^2

I also got the time from the first part which would be 0.82 seconds, but since the gravity/acceleration is different from Earth shouldn't it take longer for the jumper to go from one point to another?

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