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are 10 ft off the ground, her coworker throws a measuring tape toward her. The measuring tape is thrown from a height of 6 ft with an initial vertical velocity of \( 15 \mathrm{ft} / \mathrm{s} \). Projectile motion formula: \[ h=-16 t^{2}+v t+h_{0} \] \( t= \) time, in seconds, since the measuring tape was thrown \( h= \) height, in feet, above the ground \( h=3 t+10 \) and \( h=-16 t^{2}+15 t+6 \) \( 10 t+3 \) and \( h=-16 t^{2}+6 t+15 \) \( -16 t^{2}+3 t+10 \) and \( h=-16 t^{2}+15 t+6 \) \( -16 t^{2}+10 t+3 \) and \( h=-16 t^{2}+6 t+15 \) COMPLETE The system has no real solutions. What does this indicate about the worker's arms and the measuring tape? They will never be at the same height at the same time. They will be at the same height exactly once. They will be at the same height at two different times. DONE

Ask by Dunn Campos. in the United States
Mar 22,2025

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They will never be at the same height at the same time.

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Extra Insights

When the equations for heights of both the coworker and the measuring tape are analyzed, it becomes apparent that if there are no real solutions, it means they will never meet at the same height. This suggests that the measuring tape, despite its initial velocity, either ascends to a height that never intersects with the coworker's 10 ft or decays below her height too quickly, making the task impossible. In a real-world scenario, this situation might reflect the importance of effective communication and ensuring that tools are used in a safe and logical manner. No one wants to throw equipment that can’t reach its intended target—next time, maybe just hand over that measuring tape instead of tossing it!

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