Astone is thrown at an angle of 30° above the horizontal from the top edge of a cliff with an initial speed of 12 m/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 = 9.8 m/s2 and air resistance is negligible)

To solve this problem you must apply the proccedure shown below:
 1. You must apply the following formula:
 yo=-(Voy)t+gt²/2
 2. You have that:
 Voy=VoxSin(30°)
 Voy=(12 m/s)(Sin(30°))
 Voy=6 m/s
 t=5.6 s
 g=9.8 m/s²
 2. When you susbtitute these values into the formula, you obtain:
 yo=-(6 m/s)(5.6 s)+(9.8 m/s^2)(5.6 s)²/2
 yo=120.06 m
 Therefore, the answer is: 120.06 m