A stone is thrown at an angle of 30° above the horizontal from the top edge of a cliff with an initial speed of 8.0 m/s.A stop watch measures the stone's trajectory time from top of cliff to bottom to be 5.9 s.How far out from the cliff's edge does the stone travel horizontally? (g = 9.8 m/s² and air resistance is negligible) A)41 m B)27 m C)300 m D)130 m

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Quizplus · Accepted Answer

The horizontal distance traveled by the stone can be calculated using the formula for horizontal motion: distance = velocity * time. The horizontal component of the velocity is given by $v_x = v \cos(\theta)$, where $v = 8.0 \, \text{m/s}$ and $\theta = 30^\circ$. Thus, $v_x = 8.0 \cos(30^\circ) = 6.93 \, \text{m/s}$. The time of flight is $5.9 \, \text{s}$, so the horizontal distance is $6.93 \, \text{m/s} \times 5.9 \, \text{s} = 40.89 \, \text{m}$, which rounds to $41 \, \text{m}$.

A Stone Is Thrown at an Angle of 30° Above