Asked by Katie
A stone is thrown straight upward with a speed of 12 m/s from the edge of a 70 m high cliff. How long is the stone in the air? What is the speed just before it hits the ground? What is the total distance the stone traveled?
Answers
Answered by
drwls
(a) Height of object from cliff base
= H = 70 +12 t - (g/2) t^2
Set H = 0 and g = 9.8 m/s^2, and solve for t. Take the positive rooot.
(b) Vfinal^2 = Vinitial^2 + 2*g*70
= (144 +1372) m^2/s^2
Vfinal = 38.9 m/s
(c) (Up distance) + (Down distance)
= 2*(Up distance) + 70 m
= 2*Vo^2/(2*g)+70
= (Vo^2/g) + 70
= H = 70 +12 t - (g/2) t^2
Set H = 0 and g = 9.8 m/s^2, and solve for t. Take the positive rooot.
(b) Vfinal^2 = Vinitial^2 + 2*g*70
= (144 +1372) m^2/s^2
Vfinal = 38.9 m/s
(c) (Up distance) + (Down distance)
= 2*(Up distance) + 70 m
= 2*Vo^2/(2*g)+70
= (Vo^2/g) + 70
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