Question

A pole-vaulter is nearly motionless as he clears the bar, set 4.2 m above the ground. He then falls onto a thick pad. The top of the pad is 80 cm above the ground, and it compresses by 50 cm as he comes to rest. What is the magnitude of his acceleration as he comes to rest on the pad

Answer 1

initial height of the pole vaulter = 4.2 m

height of the pole vaulter just before it touch the pad = 80 cm

so the total displacement of pole vaulter just before he will touch the pad = 4.2 - 0.80 = 3.4 m

now by kinematics

$v_f^2 - v_i^2 = 2a d$

$v_f^2 - 0^2 - 2* 9.8*3.4$

$v_f^2 = 66.64$

$v_f = 8.16 m/s$

now after this he will come to rest after compressing the pad by 50 cm

so again we can use kinematics to find its acceleration

$v_f^2 - v_i^2 = 2 a d$

$0 - 8.16^2 = 2*a*0.50$

$a = -66.64 m/s^2$

so here its acceleration will be - 66.64 m/s^2