Question

qid 1023
a stone is thrown at an angle of 37 degrees above the horizontal from the top edge of a cliff with an initial speed of 12 m/s. a stop watch measures the stones trajectory time from top of cliff to bottom to be 8 s. what is the height (in meters) of the cliff? (air resistance is negligible).
569.43 m
57.77 m
313.6 m
255.83 m

Explanation:

Step1: Resolve the initial velocity into vertical - component

The initial vertical velocity $v_{0y}=v_0\sin\theta$, where $v_0 = 12$ m/s and $\theta = 37^{\circ}$. So $v_{0y}=12\times\sin37^{\circ}=12\times0.6 = 7.2$ m/s.

Step2: Use the kinematic equation for vertical displacement

The kinematic equation for vertical displacement is $\Delta y=v_{0y}t-\frac{1}{2}gt^{2}$, where $t = 8$ s and $g = 9.8$ m/s².
Substitute the values: $\Delta y=(7.2\times8)-\frac{1}{2}\times9.8\times8^{2}$.
First, calculate $7.2\times8 = 57.6$ and $\frac{1}{2}\times9.8\times8^{2}=4.9\times64 = 313.6$.
Then $\Delta y=57.6-313.6=- 256$ m. The negative sign just indicates the direction (down - ward). The height of the cliff is approximately $256$ m, closest to $255.83$ m considering calculation approximations.

Answer:

$255.83$ m