At impact, the force due to gravity is =
F
=
m
g
, where we see that a larger force corresponds to a larger mass. To rectify this with your intuition, as in a brick would hurt more if it landed on my head when dropped from a higher distance -- This is true (I don't recommend experiment), but the relevant quantity is
i
m
p
u
l
s
e
, which you can look up, or a more careful analysis of the acceleration of the brick when it lands on something, where we see that at contact, in a very short period of time, the velocity goes from →0
v
→
0
, so the force experienced is:
==∼ΔΔ=−0
F
=
d
p
d
t
=
m
d
v
d
t
∼
m
Δ
v
Δ
t
=
v
−
0
t
ϵ
Where we see that the faster the brick slows down (
t
ϵ
) the larger the force experienced by the brick (and equally and oppositely, your head), which I think is more in the vein of what you are asking.
