QuestionJune 10, 2025

A stone is thrown at an angle of 30^circ above the horizontal from the top edge of a cliff with an initial speed of 12m/s A stop watch measures the stone's trajectory time from top of cliff to bottom to be 5.6 s. What is the height of the cliff? (g=9.8m/s^2 and air resistance is negligible) 197 m 120 m 58 m 154 m

Solution
4.0(196 votes)

Answer

187 m Explanation 1. Calculate the vertical component of initial velocity Use v_{y0} = v_0 \sin(\theta) where v_0 = 12 \, \text{m/s} and \theta = 30^\circ. Thus, v_{y0} = 12 \sin(30^\circ) = 6 \, \text{m/s}. 2. Apply kinematic equation for vertical motion Use h = v_{y0} t + \frac{1}{2} g t^2 with t = 5.6 \, \text{s} and g = 9.8 \, \text{m/s}^2. Substitute to find h: h = 6 \times 5.6 + \frac{1}{2} \times 9.8 \times (5.6)^2. 3. Calculate height Compute h = 33.6 + 153.664 = 187.264 \, \text{m}.

Explanation

1. Calculate the vertical component of initial velocity Use $v_{y0} = v_0 \sin(\theta)$ where $v_0 = 12 \, \text{m/s}$ and $\theta = 30^\circ$. Thus, $v_{y0} = 12 \sin(30^\circ) = 6 \, \text{m/s}$. 2. Apply kinematic equation for vertical motion Use $h = v_{y0} t + \frac{1}{2} g t^2$ with $t = 5.6 \, \text{s}$ and $g = 9.8 \, \text{m/s}^2$. Substitute to find $h$: $h = 6 \times 5.6 + \frac{1}{2} \times 9.8 \times (5.6)^2$. 3. Calculate height Compute $h = 33.6 + 153.664 = 187.264 \, \text{m}$.

112

Click to rate:

Solution4.0(196 votes)

Answer

Explanation

Similar Questions

Solution
4.0(196 votes)