A stone 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
