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A rock is thrown at a window that is located 18.0 m above the ground. The rock is thrown at an angle of 40.0° above horizontal. The rock is thrown from a height of 2.00 m above the ground with a speed of 30.0 m/s and experiences no appreciable air resistance. If the rock strikes the window on its upward trajectory, from what horizontal distance from the window was it released?

A. 29.8 m
B. 27.3 m
C. 48.7 m
D. 71.6 m
E. 53.2 m

Answer :

cjmejiab

To solve the problem it is necessary to apply the kinematic equations of linear descriptive motion. With the components of speed we can find both time and distance that separates them. The two components given are:

[tex]v_x = 30cos40\°[/tex]

[tex]v_y = 30sin40\°[/tex]

Applying the vertical displacement equation we have to

[tex]y = \frac{1}{2}at^2 +v_yt +y_1[/tex]

Replacing we have,

[tex]18= \frac{1}{2}(-9.81)t^2+30sin40t+2[/tex]

Solving for t,

[tex]t = 1.189s[/tex]

Through the horizontal component we can find the displacement which is given as,

[tex]d = v_x (t)[/tex]

[tex]d = 30 cos40 (1.189)[/tex]

[tex]d = 27.324m[/tex]

Therefore the correct answer is B.

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