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

The apple will not reach the friend in the tree house as it will only reach a height of approximately 1.527 m.

Step-by-step explanation:

To determine whether the apple will reach a friend in a tree house 5.9 m above the ground, we can use the equations of motion. Since the apple is thrown vertically upward, it will experience a negative acceleration due to gravity. Using the equation h = vo*t + (1/2)*a*t^2, where h is the final height, vo is the initial velocity, a is the acceleration, and t is the time, we can calculate the time it takes for the apple to reach a height of 5.9 m. Plugging in the values, we get:

5.9 = 2.8*t + (1/2)*(-9.81)*t^2

Simplifying the equation, we have:

-4.905*t^2 + 2.8*t - 5.9 = 0

Using the quadratic formula, we can solve for t. The quadratic formula is t = (-b ± sqrt(b^2 - 4ac)) / (2a), where a = -4.905, b = 2.8, and c = -5.9.

Plugging in the values, we get:

t = (-2.8 ± sqrt(2.8^2 - 4*(-4.905)*(-5.9))) / (2*(-4.905))

After evaluating the formula, we find that the apple will take approximately 1.527 seconds to reach a height of 5.9 m. Since the apple continues to rise after reaching this height, it will not reach the friend in the tree house.

Answer 2

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

Step-by-step 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=(v^2-u^2)/(2a)\\\Rightarrow s=(0^2-2.8^2)/(2* -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.