Question

27. A stone is thrown vertically upward with an initial velocity of 40 ms Taking g=10 ms^-2. Find the
maximum height reached by the stone. What is the net displacement and the total distance covered by
stone?
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Answer 1

Maximum height covered =u^2/2g

Explanation:

solution÷

u^2/2g=40×40/2×10

=80 m

total distance covered =2×80=160 m

displacement =80 m

Answer 2

Answer:

your answer is........