Question

Stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/s^2
, find the
the maximum height reached by the stone. What are the net displacement and the total distance covered by
the stone?

Answer 1

Answer:

Maximum height is u^2/2g

= (40)^2/20

= 1600/20

= 80 m

Net displacement will be 0 because the stone will fall down below( assume that there is no rebound,even if there is rebound there is no mention of coefficient of restitution of floor so we can't calculate.)

Total distance after it falls down will be 160 m.

But the question seems ambiguous ,I mean net displacement after it reaches the height or it falls down is not mentioned.

Total distance and diplacement is same (80) if it is "after reaching max height".

Explanation: