Givens: Acceleration (a) = 32.17 ft/s² as that is the only force acting on the stone in free fall assuming no air resistance
Initial velocity (V1) = 0 ft/s as that it the starting speed of the stone as all objects about to move from rest must start at 0 velocity at rest = 0 velocity
Final velocity (V2) = 120 ft/s
Displacement aka. the height of the cliff (Δd) = ? ft
The kinematic formula we can use is: V2² = V1² + 2(a)(Δd) Isolate to solve for Δd and just plug in the numbers! V2² = V1² + 2(a)(Δd) (120)² = (0)² + 2(32.17)(Δd) 14400 = 2(32.17)(Δd) [tex] \frac{1400}{2(32.17)} [/tex] = Δd 223.9 = Δd
Therefore, the height of the cliff is 223.9 ft! Hope this helped! :)
