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

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.

vi^2=2g*height solve for max height.

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

how long will it be in the air before it hits the ground?

To determine whether the apple will reach the tree house, we need to calculate the maximum height the apple will reach and compare it to the height of the tree house.

We can use the equation for vertical motion to calculate the maximum height (h) the apple will reach:
v_f^2 = v_i^2 + 2aΔy

Given:
Initial velocity (v_i) = +2.7 m/s (upward)
Final velocity (v_f) = 0 m/s (at the highest point, the apple momentarily stops)
Acceleration (a) = -9.8 m/s^2 (due to gravity, acting downward)
Change in position (Δy) = Maximum height (h) - Initial height (1.2 m) = h - 1.2 m

Plugging these values into the equation, we have:
0^2 = (2.7)^2 + 2(-9.8)(h - 1.2)

Simplifying the equation:
0 = 7.29 - 19.6h + 23.52
19.6h = 30.81
h = 30.81 / 19.6

Calculating the maximum height:
h ≈ 1.57 m

Now, we can compare the maximum height (1.57 m) to the height of the tree house (4.8 m).

Since the maximum height the apple reaches is less than the height of the tree house, the apple will not reach the tree house. Therefore, the correct answer is option 3: No, the apple will reach 3.22844 m below the tree house.