A 57.6 kg pole-vaulter falls from rest from a height of 5.4 m onto a foam-rubber pad. The pole-vaulter comes to rest 0.34 s after landing on the pad.

Question

(a) Calculate the athlete's velocity just before reaching the pad.



(b) Calculate the constant force exerted on the pole-vaulter due to the collision

Answer 1

First realize that this is a free fall problem. Meaning, you are dealing with "y direction" motion where you have constant acceleration of 9.8 m/s2 acting downwards, towards the earth. Next, you assume which direction you want to take positive. Let's say, upwards is positive.

Given

v_i = 0 since pole-vaulter falls from rest

Δy = -5.4 m (because upwards is positive, and s/he is going downwards)

acceleration, a = -g = -9.8 m/s² (again, negeative because acceleration acting downwards, and you assumed positive direction upwards)

Part a) From kinematics

v_f² = v_i² + 2a(Δy) = 0 + 2 *(-9.8 m/s²)* (-5.4 m) = 105.84 m²/s²

v_f = √(105.84) = 10.287 m/s

Part b) The v_f from part a) is what you have as initial velocity of the next phase in determining force which is hitting the floor and slowing down to rest. So, in this phase, your v_i = -10.287 m/s (direction downwards, right?) and v_f = 0 as s/he comes to rest. The slowing down takes place in 0.34 sec which is given in the problem. Don't confuse this time of 0.34 sec with the time to freefall which we haven't (and did not need either!) calculated.

Now,

F = m * a

F = m * Δv/Δt --> definition of ave acceleration

F = m * (v_f - v_i)/ Δt

F = m * (0 - (-10.287) )/ Δt

F = m * 10.287/Δt

F = 57.6 kg * 10.287/ 0.34 sec

F = 1742. 88 N

(Notice postive value for Force. So, according to your sign selected convention, it's acting upwards)