Air resistance of pine cone

[get_rekd]

A 0.25 kg pine cone falls from a branch 20 m above the ground.

A) With what speed would it hit the ground if air resistance could be ignored?
m= 0.25 kg
g= 9.8 m/s^2
d= 20 m

Ep= (0.25kg)(9.8m/s^2)(20m)
= 49 J
Ek= 1/2mv^2
49J = 1/2(0.25kg)(v^2)
2(49 J = (0.5kg)(0.5 v^2))
98 J = (0.5 kg)(0.5 v^2)
-0.5 -0.5
97.5 J = 0.5 v^2
----------------
0.5 0.5
v^2 = 195
v = 14 m/s
can someone check this for me please?

part b)
If the pine cone actually hits the ground with a speed of 9.0 m/s, what was the average force of air resistance on it?

I am not sure how to solve this problem?
Can someone please walk me through it?

 

[mgb_phys]

For part a I get, using mgh=1/2mv^2, v=sqrt(2gh) = sqrt(2*20*9.8) = 19.8m/s

For part B. Work out the KE it hits the ground with, the KE it should have had, and so how much E is lost ot friction.
Then from Energy = force * distance, you can work out an average force.

 

[get_rekd]

is mgh=1/2mv^2=sqrt(2*20*9.8) a shortcut or is that the way that I should setup the problem from the start? What is the 2 for in the equation. I understand that 20 = h and 9.8 = g

 

[get_rekd]

anyone?

 

[mgb_phys]

Kinetic energy = 1/2 m v^2.
Potential energy = m g h
If the object falls from rest (no initial velocity) and no energy is lost to friction, then the easiest way of working out the speed is to assume all the potential energy is converted to kinetic energy. Note that the mass of the object cancels out - the speed something falls at doesn't depend on it's mass.