Asked by helpless
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
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
Answers
Answered by
bobpursley
vi^2=2g*height solve for max height.
Answered by
Damon
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
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
Answered by
clue
how long will it be in the air before it hits the ground?
Answered by
Henry
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.
(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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