Posted by **helpless** on Friday, November 12, 2010 at 8:11pm.

Maria throws an apple vertically upward from a height of 1.2 m with an initial velocity of +2.7 m/s.

Will the apple reach a friend in a tree house 4.8 m above the ground?

the tree house

2. No, the apple will reach 1.40733 m below

the tree house

3. No, the apple will reach 3.22844 m below

the tree house

4. No, the apple will reach 1.27691 m below

the tree house

5. Yes, the apple will reach 3.22844 m above

the tree house

6. Yes, the apple will reach 1.40733 m above

the tree house

- physics -
**bobpursley**, Friday, November 12, 2010 at 8:23pm
vi^2=2g*height solve for max height.

- physics -
**Damon**, Friday, November 12, 2010 at 8:26pm
v = vi - 9.8 t

v = 0 at max height

so

t = 2.7/9.8 time to top = .276

h = 1.2 + 2.7 t -4.9 t^2

=1.57

4.8 - 1.57 = 3.22 below the house

- physics -
**clue**, Friday, November 12, 2010 at 8:57pm
how long will it be in the air before it hits the ground?

- physics -
**Henry**, Saturday, November 13, 2010 at 11:41pm
Vf^2 = Vo^2 + 2gd = 0,

(2.7)^2 + 2(-9.8)d = 0,

7.29 - 19.6d = 0,

-19.6d = - 7.29,

d = -7.29 / -19.6 = 0.372 m. upward.

d(tot.) = 1.2 + 0.372 = 1.57 m.

4.8 - 1.57 = 3.228 m below the house.

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