How high is the tree branch(conservation in energy/kinematics)

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Homework Statement


45a)
A watermelon with a mass of 2 kg falls out of a tree house that is 5.4 m above the ground. What is the speed of the watermelon just before it hits the ground.
(this part i have solved)

mass = 2kg
Δd= 5.4m
v=?

v1^2 =v2^2+2gΔd

v1^2=0+2gΔd

=0+2(9.8)(5.4)

v1^2=√105.84

v1^2=10.28 m/s


Now on to the part i didnt quite get.

45b) A cantaloupe with a mass of 0.45kg falls out of the other side of the tree house. It hits a tree branch at a speed of 6.3 m/s. How high is the tree branch from the ground?


Homework Equations



v1^2 =v2^2+2gΔd


v1^2-v2^2/2g=Δd


The Attempt at a Solution





45b)

v1=0m/s

v2=6.3m/s

mass = 0.45 kg

v1^2-v2^2/2g=Δd

v1^2-v2^2/2g=Δd

6.3^2-0/2g=Δd


6.3^2/2(9.8)=Δd


6.3^2/19.6=Δd


39.69/19.6=Δd


2.025m = Δd

thats the height i got, if it is not correct, can someone show me how to get the right height? or better yet get the right height and i can analyze and figure out how you got the right height?
 
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Discussion

Hi supernova1203. That would be the height from the branch to the treehouse, the question asks for the height of the branch from the ground.
 
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Sleepy_time said:
Hi supernova1203. That would be the height from the branch to the treehouse, the question asks for the height of the branch from the ground.


Tell me how did you come to that conclusion?

ah...so in order to get the height from the branch to the ground, i would have to subtract the 2 heights...max height and the 2nd height i got for part B of the question?
 
" A cantaloupe with a mass of 0.45kg falls out of the other side of the tree house. " The tree house is 5.4m above the ground from a), so I think by other side of the tree house just means it is dropped from the same height as in a). And yes that is the answer