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Physics
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in a movie, a monster climbs to the top of a building 40 meters above the group and hurls a boulder downward with a speed of 28 m/s at an angle of 48 degrees below the horizontal. how far from the building does the boulder land?  

Oct 1st, 2014

Coordinate system: origin at the initial position of the boulder, x-axis directed to the right, y-axis directed upwards.

Given: h = 40 m, v_0 = 28 m/s, theta = - 48°. Find x_f (final). We know that y_f = - 40.

 Equations of the motion: x = v_0, x * t; y = v_0, y * t - g t^2/ 2, 

where v_0, x = v_0 * cos(theta) = 28 * cos(-48°) = 18.7 m/s and 

v_0, y = v_0 * sin(theta) = 28 * sin(-48°) = -20.8 m/s.

For the final time: - h = v_0, y * t - g*t^2/2 we have a quadratic equation g*t^2/2 - v_0, y * t + h = 0 whose solutions are t = 2*(v_0, y +/- sqrt{(v_0, y)^2 + 2*g*h})/g = 2*(-20.8 +/- sqrt{20.8^2 + 2*9.81*40}) = 1.47 s.

In the last formula we took only plus sign since the solution with minus sign is negative and does not make sense. Finally, x = v_0, x * t = 18.7 * 1.47 = 27.5 m.

Answer: The boulder will land 27.5 m away from the building.


 

Mar 18th, 2015

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