A stone is allowed to fall from the top of a tower 100 m high and at the same time another stone is projected vetically upwards from the ground with a velocity of 25 m/s. Calculate when and where the two stones will meet.
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Answer:
Let the stones meet at point A after time t.
For upper stone :
u
′
=0
x=0+
2
1
gt
2
x=
2
1
×10×t
2
⟹x=5t
2
............(1)
For lower stone :
u=25 m/s
100−x=ut−
2
1
gt
2
100−x=(25)t−
2
1
×10×t
2
⟹100−x=25t−5t
2
............(2)
Adding (1) and (2), we get
25t=100
⟹t=4 s
From (1),
x=5×4
2
⟹x=80 m
Hence the stone meet at a height of 20 m above the ground after 4 seconds.
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