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Maria throws an apple vertically upward from a height of 1.4 m with an initial velocity of +3.1 m/s. Will the apple reach a friend in a tree house 4.2 m above the ground? The acceleration due to gravity is 9.81 m/s 2 .

Respuesta :

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According to the problem, you are trying to find a height. If the height a the top of the path is greater than or equal to 4.2m, then the apple will reach Maria's friend. If the height is less than 4.2m, the apple will not reach maria's friend. You are given the known quantities: initial velocity, acceleration, and you can set final velocity = 0. This means you can use the kinematic equation:
[tex]v_f^2 = v_i^2 + 2a\Delta x[/tex]

Explanation:

It is given that,

Initial position of the Apple, [tex]h_o=1.4\ m[/tex]

Initial velocity, u = 3.1 m/s

Position of friend, Y = 4.2 m

Using second equation of motion as :

[tex]Y=ut-\dfrac{1}{2}gt^2+h_o[/tex]

[tex]4.2=3.1t-\dfrac{1}{2}\times 9.8t^2+1.4[/tex]    

[tex]3.1t-4.9t^2-2.8=0[/tex]..............(1)

On solving the above quadratic equation we get t = 2.02 seconds

Since, the value of t is real. So, the apple will reach a friend in a tree house. Hence, this is the required solution.                                    

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