A rescue plane spots a survivor 132 m directly below and releases an emergency kit with a parachute.If the package descends at a constant vertical acceleration of 6.89 m/s² and the initial plane horizontal speed was 68.9 m/s,how far away from the survivor will it hit the waves? A)426 m B)2.64 km C)301 m D)446 m

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Quizplus · Accepted Answer

The horizontal distance traveled by the package is equal to the initial horizontal speed multiplied by the time it takes to fall 132 m. The time it takes to fall 132 m can be found using the equation of motion:$$y_f = y_i + v_i t + \frac{1}{2} a t^2$$where:* $y_f$ is the final vertical position (132 m)* $y_i$ is the initial vertical position (0 m)* $v_i$ is the initial vertical velocity (0 m/s)* $a$ is the vertical acceleration (-6.89 m/s^2)* $t$ is the time (in seconds)Solving for $t$, we get:$$t = \sqrt{\frac{2(y_f - y_i)}{a}} = \sqrt{\frac{2(132 	ext{ m})}{6.89 	ext{ m/s}^2}} = 5.64 	ext{ s}$$Therefore, the horizontal distance traveled by the package is:$$x = v_i t = 68.9 	ext{ m/s} 	imes 5.64 	ext{ s} = 426 	ext{ m}$$

A Rescue Plane Spots a Survivor 132 M Directly Below