1. ## free falling.

A student (m = 63 kg) falls freely from rest and strikes the ground.

During the collision with the ground, he comes to rest in a time of 0.0200 s.

The average force exerted on him by the ground is +18500. N, where the upward direction is taken to be the positive direction.

From what height did the student fall? Assume that the only force acting on him during the collision is that due to the ground.
in meters

is this a problem solved using a kinematic equation?

2. Originally Posted by rcmango
A student (m = 63 kg) falls freely from rest and strikes the ground.

During the collision with the ground, he comes to rest in a time of 0.0200 s.

The average force exerted on him by the ground is +18500. N, where the upward direction is taken to be the positive direction.

From what height did the student fall? Assume that the only force acting on him during the collision is that due to the ground.
in meters

is this a problem solved using a kinematic equation?
No, but that is a part of it.

In order to use a kinematic equation, or an energy equation for that matter, we need to know how fast the student was falling just before impact. We can get that number by using the Impulse-Momentum Theorem. As the problem stated, I'm going to use upward as a +y axis.

$J = \Delta p$

$\bar{F} \Delta t = p - p_0$

$\bar{F} \Delta r = m(v - v_0)$

The final speed of the student is 0 m/s, so
$\bar{F} \Delta t = -mv_0$

So
$v_0 = - \frac{\bar{F} \Delta t}{m}$

So: $\bar{F} = 18500~N$, $\Delta t = 0.0200~s$, and $m = 63~kg$.

Note that your $v_0$ will be negative. This is because + is upward and the velocity just before impact is downward. (This is a good check to see if our equation is right.)

Once you get $v_0$ you can either do a work-energy or kinematics problem to determine the height.

-Dan

3. So, even though the problem says it starts from rest, I must assume the v0 is not 0, because it must have an initial JUST before it hits the ground which would be the initial velocity?

...okay so, v0 is -5.87

so i used a kinematics equation of: d = ( (vi + vf) / 2) * t

where d is:

d =vi * t + 1/2 * a * t^2

d = -5.87 + 1/2* (9.8)* .0200^2

...and that did not work.

so i found t by using vf + vi/ a .... 0 + 5.87 / 9.8 = t

t = .598979592

okay so i tried to

.....................nevermind its 1.76 meters =D

i just used a different kinematics equation.

4. Originally Posted by rcmango
So, even though the problem says it starts from rest, I must assume the v0 is not 0, because it must have an initial JUST before it hits the ground which would be the initial velocity?

...okay so, v0 is -5.87

so i used a kinematics equation of: d = ( (vi + vf) / 2) * t

where d is:

d =vi * t + 1/2 * a * t^2

d = -5.87 + 1/2* (9.8)* .0200^2

...and that did not work.

so i found t by using vf + vi/ a .... 0 + 5.87 / 9.8 = t

t = .598979592

okay so i tried to

.....................nevermind its 1.76 meters =D

i just used a different kinematics equation.
This post sounds like you are using any equation you can think of to find an answer, even if it doesn't apply, and if it doesn't work you are moving on to the next one.

You need to be careful about the definitions. Perhaps I erred when I used $v_0$ to represent the speed of the student just before collision. This $v_0$ is not the same as the initial velocity with which the student started to fall.

We already have +y upward. I'm going to put an origin on the ground. So let's fill out a table.
$t_0 = 0~s$
$y_0 =$? and $y = 0~ m$
$v_0 = 0~m/s$ and $v = -5.87302m.s$
(This final velocity is the $v_0$ from the impulse-momentum part of the problem.)
$a = -g$

We are looking for $y_0$. Given the known and unknown information you can immediately see that
$v^2 = v0^2 + 2a(y - y_0)$
will do the trick.

Plan out how you are going to do the problem, then organize it.

-Dan