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

-67.24 m/s²

Explanation:

From the given information:

Using the formula for Kinematic equation i,e to find the speed with which the pole-vaulter approaches the pad

where;

the speed when the individual approaches the pad = ???

initial speed at the location of the bar = 0

the individual acceleration which is also equivalent to acceleration due to gravity. = 9.8 m/s²

d = distance travelled

Thus;

d = 4.2m - 80 cm

d = 4.2 m - 0.8 m

d = 3.4 m

Thus, the initial speed of the pole-vaulter at the top of the pad = 8.2 m/s

Using the same above equation to determine the acceleration as the pole-vaulter approaches rest on the pad:

i.e

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j