an object is thrown off a cliff with intial pward velocity of 15m/s it strikes the ground with downward velocity of 25 m/s what is the height of the cliff

20 m

Explanation:

Given:

v₀ = 15 m/s

v = -25 m/s

a = -10 m/s²

Find: Δy

v² = v₀² + 2aΔy

(-25 m/s)² = (15 m/s)² + 2 (-10 m/s²) Δy

Δy = 20 m

the electric field is a “force field” around a charged object that illustrates the direction the electric force would push an imaginary