A stone is dropped from the top of a tower 400 m high and at the same time another stone is projected upward
vertically from the ground with a velocity of 100 ms −1 . Find where and when the two stones will meet.
3 answers
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S=1/2at^2
400-s=ut-1/2at^2
400-5t^2=100t-5t^2
t=4s
S=5t^2
S=80m from above
or 400-80=320m from below
400-s=ut-1/2at^2
400-5t^2=100t-5t^2
t=4s
S=5t^2
S=80m from above
or 400-80=320m from below
x=Vot+1/2at^2
For downward motion Vo=0,
g=9.8m/s
h=Vot+1/2at^2
=0+1/2×9.8t^2
=4.9t^2
Forupward motion Vo=100
g=-9.8m/s
|||^ly
t=4s
From above,
h=4.9t^2
=4.9 (4)^2
=4.9(16)
78.4m
Therefore,
The two stones meet at 78.4m below the tp of the tower after 4seconds.
For downward motion Vo=0,
g=9.8m/s
h=Vot+1/2at^2
=0+1/2×9.8t^2
=4.9t^2
Forupward motion Vo=100
g=-9.8m/s
|||^ly
t=4s
From above,
h=4.9t^2
=4.9 (4)^2
=4.9(16)
78.4m
Therefore,
The two stones meet at 78.4m below the tp of the tower after 4seconds.
4 answers