A stone is thrown vertically upwards from the
ground and caught at the same point on the
ground after 10 seconds. (Take, g = 10 m/s2)
The maximum height reached by stone is
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Answer:
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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