Question

Maria throws an apple vertically upward from a height of 1.3 m with an initial velocity of +2.8 m/s. Will the apple reach a friend in a tree house 5.9 m above the ground? The acceleration due to gravity is 9.81 m/s 21. No, the apple will reach 5.27136 m below the tree house2. Yes, the apple will reach 5.27136 m above the tree house3. No, the apple will reach 1.43117 m below the tree house4. No, the apple will reach 1.5289 m below5. Yes, the apple will reach 1.5289 m above the tree house6. Yes, the apple will reach 1.43117 m above the tree house

Answer 1

Answer:

No, the apple will reach 4.20041 m below the tree house.

Explanation:

t = Time taken

u = Initial velocity = 2.8 m/s

v = Final velocity = 0

s = Displacement

g = Acceleration due to gravity = -9.81 m/s² = a (negative as it is going up)

Equation of motion

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-2.8^2}{2\times -9.81}\\\Rightarrow s=0.39959\ m$

The height to which the apple above the point of release will reach is 0.39959 m

From the ground the distance will be 1.3+0.39959 = 1.69959 m

Distance from the tree house = 5.9-1.69959 = 4.20041 m

No, the apple will reach 4.20041 m below the tree house.

The values in the option do not reflect the answer.