Solution to Example from Module 1

Solution to Example from Module 1:

A stone is dropped from a 70-meter cliff. Neglect air resistance. a) What is the velocity of the stone after 2 seconds?

Known: Up will be considered positive and down will be considered negative a = g = -9.81 m/s2 t = 2 s vi = 0 m/s (because it was dropped we assume it was initially at rest) vf = ?

Solution: vf = vi + at 2 vf = 0 m/s + (-9.81 m/s )(2 s) vf = 0 m/s + (-19.62 m/s) vf = -19.62 m/s

After 2 seconds the stone is falling downwards at a speed of 19.62 m/s. b) When does the stone hit the ground?

Known: Up will be considered positive and down will be considered negative. a = g = -9.81 m/s2 d = -70 m (again, the distance is negative because the stone moved downward) vi = 0 m/s t = ?

Solution: 1 2 d = vit + /2at 1 2 2 -70 m = (0 m/s)(t) + /2(-9.81 m/s )(t ) -70 m = 0 + (-4.905 m/s2)(t2) t2 = (-70 m)/(-4.905 m/s2) t2 = 14.27 s2 t = 3.8 s

The stone will hit the ground after 3.8 seconds.