Question

A pole-vaulter of mass 60.0 kg vaults to a height of 6.0 m before dropping to thick padding placed below to cushion her fall.

(a) Find the speed with which she lands.

(b) If the padding brings her to a stop in a time of 0.50 seconds, what is the average force on her body due to the padding during that time interval?

Answer 1

Answer

given,

mass of the pole-vaulter = 60 Kg

height of the vaults = 6 m

acceleration due to gravity = g = 9.8 m/s²

a) Speed of landing

     $v = \sqrt{2gh}$

     $v = \sqrt{2\times 9.8 \times 6}$

     $v = \sqrt{117.6}$        

     $v = 10.84\ m/s$  

b) time to stop = 0.50 s

   impulse = F t

   impulse = mv - mu

   Ft = m(v - u)

   $F = \dfrac{m(v-u)}{t}$

   $F = \dfrac{60(0 - 10.84)}{0.5}$

   $F = 1300\ N$