You throw an 18.0 N rock into the air from ground level and observe that, when it is 15.0 m high, it is traveling upward at 17.0 m/s. Part A Use the work-energy principle to find the rock's speed just as it left the ground. Express your answer in meters per second. vground

Vi = 24.14 m/s

Explanation:

If we apply Law of Conservation of Energy or Work-Energy Principle here, we get: (neglecting friction)

Loss in K.E of the Rock = Gain in P.E of the Rock

(1/2)(m)(Vi² - Vf²) = mgh

Vi² - Vf² = 2gh

Vi² = Vf² + 2gh

Vi = √(Vf² + 2gh)

where,

Vi = Rock's Speed as it left the ground = ?

Vf = Final Speed = 17 m/s

g = 9.8 m/s²

h = height of rock = 15 m

Therefore,

Vi = √[(17 m/s)² + 2(9.8 m/s²)(15 m)]

Vi = √583 m²/s²

Vi = 24.14 m/s