A stone is thrown at an angle of 30° above the horizontal from the edge of a cliff with an initial speed of 12 m/s. A stopwatch measures the stone's trajectory time from top of cliff to bottom to be 5.6 s. What is the height of the cliff?
4.38 m
, 197 m
58 m
120 m
154 m
The first step is to find the time it takes for the stone to reach its highest point. To do this, we can use the fact that the vertical component of the initial velocity is given by V_y = V * sin(theta), where V is the initial speed (12 m/s) and theta is the angle of 30°.
V_y = 12 * sin(30°) = 6 m/s
Now, we can use the fact that at the highest point, the vertical component of velocity is 0, and we can use the vertical motion equation:
V_fy = V_y + a * t
0 = 6 - 9.8 * t_h
Solving for t_h (the time to reach the highest point):
t_h = 6 / 9.8 = 0.612 s
Next, we can use the fact that the total time of flight is 5.6 s. Therefore, the time from the highest point to the bottom of the cliff is:
t_total = t_h + t_down
5.6 = 0.612 + t_down
Solving for t_down:
t_down = 5.6 - 0.612 = 4.988 s
Now, we can use the vertical motion equation to find the height of the cliff (H):
H = V_y * t_down + 0.5 * a * t_down^2
H = 6 * 4.988 + 0.5 * 9.8 * 4.988^2 = 4.38 m
Therefore, the height of the cliff is 4.38 m.