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

i don't understand. explain it clearly

one person pulls 90n one way and the other has 100n pulling the other way.

so the net force = 90 - 100 = -10n,

or the ans can be 10n if the picture shows -90n and 100n instead.