1.

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

Answer»

Solution :h = s
`V^(2) - U^(2) = `2gs
`0^(2) - (40)^(2) = 2 xx (-10) `s
s = 80 m
Total distance = 80 m + 80 m
= 160 m


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