# Projectile motion

1. Feb 27, 2008

### aLiase

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

d = v1 + at^2 / 2
v = d/t

3. 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?

Last edited: Feb 27, 2008
2. Feb 27, 2008

### blochwave

You're missing a t in your equation for d

So he.....jumps in a straight line, skidding along the ground? I believe they call that running(harhar *crickets chirp*)The angle for that would be 0, so I'm not sure how you found 22 degrees from that, unless you did two different approaches, in which case your answers contradict

And then for some reason on the moon you switched it and he jumps straight up. Although I think that's a typo, it's still the same issue with the other case

3. Feb 27, 2008

### aLiase

Ok, this is how I did it step by step.

First I found the time it toke from the ground to the maximum height using:

Vertical direction:
d = v1 + a(t)^2 / 2
0.82 = 0 + 9.81(t)^2 / 2
0.82 / 4.905 = t^2
t = 0.41
t(total) = 0.82
Then I multiply that by 2 to get the time it takes for him to jump from one point to the other.

Now I get the initial velocity since I have time I do:
Horizontal direction:
v = d / t
v = 8.12 / 0.82
v = 10 m/s

Then I used the equation v(f)^2 = v(i)^2 + 2ad to find the final velocity in the vertical direction:
v(f)^2 = 0 + 2(9.81)(0.84)
v(f) = sqrt(2(9.81)(0.84))
v(f) = 4 m/s

Then I used Tan(theta) = vertical / horizontal:
theta = tan^-1 (4 / 10/)
theta = 22 degrees

And that how I got the 22 degrees

4. Feb 27, 2008

### blochwave

Why are you assuming the initial velocity in the y direction is 0? He would never leave the ground

The reason you get an answer that kind of makes sense is because you forgot to include the negative when dealing with his vertical acceleration from gravity, so you're saying he starts with no initial vertical velocity, yet has a final vertical velocity of 4 m/s

So he falls up?