18. A stone is thrown vertically upward with an initial velocity of 40 ms-l. Taking g = 10 ms-2, draw the velocity-time graph of the motion of stone till it reaches back the ground. Use this graph to find the maximum height reached by the stone. What is the net displacement and total distance covered by the stone ?
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According to the equation of the motion under gravity
v
2
−u
2
=2gs
u=initial velocity of the stone=40m/s
v= Final velocity of the stone=0m/s
Let h be the maximum height attained by the stone
Therefore,
0
2
−40
2
=2(−10)h
h=(40×40)/20=80
Therefore, total distance covered by the stone during its upward and downward journey=80+80=160m
Net displacement during its upward and downward journey=80+(−80)=0
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