Question

A 55kg arrow is fired vertically upward at a speed of 42m/s. It rises upwards, then drops and lands on the ground. If it lodges itself 12cm in the ground, determine the average force that the ground exerts on the arrow

This is what I've done but I don't think it's correct:

Eg=mgh
=5.5x9.8x15
=808.5J

Ek=1/2(5.5)(42)
=115.5J

Em=Eg + Ek

At - 0.12m (when the arrow is in the ground:

Ek=0 therefore Eg must equal 924J due to the law of conservation of mass

Answer 1

the work going into the ground is average force x distance. That has to equal the initial energy of the arrow.
Initial energy=InitKE+InitPE
so we ignore the initial potential energy, as it is minimal.
initial KE=1/2 m v^2
setting final work=initial energy
ForceAvg*.12=1/2 *55*42^2
forcavg=(1/.24)*42^2*55
That is quite a force, not likely. However, your mass of 55kg is quite unlikely also, unless you are shooting a sixth grader of 55kg into the air. Recheck your data.