A ball is dropped from a 19m high cliff. The acceleration on the ball was 9.8m/s². What was the ball's final velocity before hitting the ground?

19.3 m/s

Explanation:

Take down to be positive.  Given:

Δy = 19 m

v₀ = 0 m/s

a = 9.8 m/s²

Find: v

v² = v₀² + 2aΔy

v² = (0 m/s)² + 2 (9.8 m/s²) (19 m)

v = 19.3 m/s

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j