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

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

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.

grav force prop 1/r squared 2r all squared is 4rsquared 1/4rsquared quartered

u

explanation: