Asked by clariza gonzalez

A 4.20-kg watermelon is dropped from rest from the roof of a 28.0-m -tall building and feels no appreciable air resistance. Calculate the work done by gravity on the watermelon during its displacement from the roof to the ground. Just before it strikes the ground, what is the watermelon's kinetic energy? Just before it strikes the ground, what is the watermelon's speed? Would the answer in part A be different if there were appreciable air resistance? Would the answer in part B be different if there were appreciable air resistance? Would the answer in part C be different if there were appreciable air resistance?
Answered by Elena
(a)W(gravity) = mgh

(b) mgh =mv²/2
v=sqrt(2gh)
(c) the same
(d)
mgh = W(fr) + mv1²/2
v1 <v
Answered by Henry
(d) Just before it strikes the ground, what is the watermelon's speed?

Vterminal = sqrt(2*acceleration*displacement) = sqrt(2*9.8 m/s^2*20 m) = 19.80 m/s
Answered by LAura
a) W=m*g*h
W=(4.20kg)(28.0m)(9.8m/s^2)=1152J
b)K=1/2*m*v^2 & v=sqrt2*g*h
v=sqrt(2)(9.8m/s^2)(28.0m)=23.4 m/s
K=(1/2)(28.0m)(23.4)^2=7665.8J
c)23.4m/s
d)no.
e)yes. Air resistance would do negative work
f)yes. Air resistance would do negative work
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