00:01
So let's draw the free body diagram for the man.
00:03
Going up would be the force subp or the fourth of the push.
00:09
Going down would be the man's weight.
00:12
So we know that the velocity that the person must have when losing contact with the ground is found from equation 212c.
00:21
We can use the acceleration due to gravity.
00:24
And then we're going to simply solve for the initial acceleration, knowing that the final acceleration is zero.
00:30
So we can say v final squared equals the initial squared plus 2g times delta x.
00:39
And here squared is going to be zero.
00:44
Therefore, v initial would be equal to the square root of negative 2g delta x.
00:49
Let's solve.
00:51
So this would be the square root of negative 2 times negative 9 .80 meters per second squared.
00:57
Again then multiplied by delta x of 0 .80 meters and we find that v initial is going to be equal to 3 .960 meters per second.
01:10
Now at this point we know that this velocity is the velocity that the jumper must have as a result of pushing with their legs...