Question
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.
vertically from the ground with a velocity of 100 ms −1 . Find where and when the two stones will meet.
Answers
Writeacher
Be sure to read through the Related Questions below. Many of these questions are very similar, so when you find a formula for your particular assignment, use it.
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.
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