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a) A stone is dropped from a height 100 m and at the same time another one in projected vertically
upwands with a velocity of 100m/sec. find when & Where the two stones meer?
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Answer:
the stone meet at a height of 20 m above the ground after 4 seconds.
Explanation:
Let the stones meet at point A after time t.
For upper stone :
u′=0
x=0+21gt2
x=21×10×t2
⟹x=5t2 ............(1)
For lower stone :
u=25 m/s
100−x=ut−21gt2
100−x=(25)t−21×10×t2
⟹100−x=25t−5t2 ............(2)
Adding (1) and (2), we get
25t=100
⟹t=4 s
From (1),
x=5×42
⟹x=80 m
Hence the stone meet at a height of 20 m above the ground after 4 seconds.
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