JUMPING FROM A HEIGHT

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics Physics 8.01X Fall Term 2002 JUMPING FROM A HEIGHT

h  ∆h                 t=0 t=t1 t=t2 In this example we will consider what happens if you bend your knees when you hit the ground if you are jumping from a height. Imagine you have mass m and jump from height h at time t = 0. You hit the ground at time t = t1 . Then between t = t1 and t = t2 over interval ∆t = t2 − t1 you bend your knees and lower your center of mass by a distance ∆h. In the figure the red spot indicates your center of mass. We will ask the questions: a. What is the average force of the ground on your legs during the impact, in terms of m, g, h and ∆h? b. What is ∆t over which the impact happens, in terms of g, h and ∆h? c. If your mass is m = 60 kg, what is the maximum h/∆h that you can sustain without breaking your tibia? We will assume that the compressive force per area necessary to break the tibia in the lower leg is about 1.6×103 bars (1 bar = 105 Pa = 105 N/m2 ). The smallest cross-sectional area of the tibia, about 3.2 cm2 , is slightly above the ankle. For this example, we’ll consider you to be a point particle located at your center of mass (we’ll see a bit later in the class why this is reasonable). To solve part (a), first let’s consider the first time interval from t = 0 to t = t1 , during which you are falling. You can get your final velocity at the end of this interval from either 1D kinematics or the work-energy theorem. Let’s do the latter. We’ll assume no friction, so there’s no non-conservative work, so ∆KE + ∆P E = 0 1 2 mv − mgh = 0 2 1 so v1 = . q (1) 2gh Now let’s consider the time interval from t = t1 to t = t2 , during the impact. We can apply the work-energy theorem again, ∆KE + ∆P E = WN C But this time, there is non-conservative work being done on you! There is a contact force from the floor on you, and its direction is antiparallel to your displacement. So the contact force does negative work on you, and this non-conservative work is WN C = −Ff loor ∆h. (We’ll assume this force is constant over the interval). The work-energy theorem for this interval gives: 1 − mv12 − mg∆h = −Ff loor ∆h 2 (where the “∆” refers to before and after the impact. Before, you have velocity v1 . After, you have come to a stop and you have velocity zero. Your change in P.E. over the interval is −mg∆h.) Now we can plug in 21 mv12 = mgh from equation1, and rearrange terms a little to get Ff loor = mgh+mg∆h ∆h So that’s part (a). Now for part (b), let’s find the impulse in order to find ∆t, since impulse is average total force times time. Impulse is also change in momentum, ∆p. We know that your change in momentum over the interval is ∆p = p2 − p1 = 0 − mv1 = −mv1 . ∆p = Ftot ∆t (which is another way of stating Newton’s 2nd Law), where Ftot is the total average force on you. There are two forces acting on you: gravity and the force of the floor. So Ftot = Ff loor − mg, where Ff loor is the average force of the floor that we just calculated in part (a). So, the magnitude of ∆t is then |F|∆p| tot | Plugging in: ∆t = |∆p| = |Ff loor − mg| mv1 mgh+mg∆h − ∆h mg The m’s cancel, and plugging in for v1 and massaging a little more we get: ∆t = √ 2gh g(h/∆h) For part (c), we use our expression for Ff loor to get Ff loor à h +1 = mg ∆h ! The maximum force that the smallest area of the tibia can take is 1.6×103 bars times 105 N/m2 per bar times 3.2 × 10−4 m2 times 2 (for two legs) is Fmax = 1.0 × 105 N. If we take Fmax = Ff loor when the tibia just breaks, and solving for h/∆h, we get (h/∆h)max = 173 . So, if you don’t bend your knees (take ∆h = 1 cm), you will break your legs jumping from only 1.7 m. If you bend your knees 0.5 m, your leg bones may survive a leap from 87 m! (Please don’t try this yourself though!! This problem considers only damage to bones– in fact other tissues in your body could get damaged in a fall from a height of more than a few meters). And if you are falling into something soft and cushiony, or into water, ∆h (and ∆t) are relatively larger. Parachutists are trained to maximize time and displacement of impact when landing by crouching and rolling. And compare a dive to a belly-flop: small ∆h and ∆t during the collision ⇒ hurts more!