A stone is thrown vertically upward with an initial velocity of 40m/s. Taking g=10m/s2, find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?
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Solution
Given : Initial velocity of stone u=40m/s
Let the maximum height reached and time taken to reach that height be H and t respectively.
Velocity of the stone at maximum height is zero i.e v=0
Using, v2−u2=2aH where a=−g=−10m/s2
0−(40)2=2(−10)H⟹H=80 m
The stone, after reaching the maximum height, starts to fall down and reach the surface again.
∴ Net displacement of the stone is Zero
Total distance covered by the stone S=2H=2×80=160 m