A stone is allowed to fall from the top of a tower 10 m high and at the same time another stone is projected vertically upward from the ground with a velocity of 25 m/s. Calculate when and where the two stones will meet. (Hint. Equations of motion)
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Explanation:
Solution
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Let the stones meet at point A after time
t
.
For upper stone :
u
′
=
0
x
=
0
+
1
2
g
t
2
x
=
1
2
×
10
×
t
2
⟹
x
=
5
t
2
............(1)
For lower stone :
u
=
25
m
/
s
100
−
x
=
u
t
−
1
2
g
t
2
100
−
x
=
(
25
)
t
−
1
2
×
10
×
t
2
⟹
100
−
x
=
25
t
−
5
t
2
............(2)
Adding (1) and (2), we get
25
t
=
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.
Answer:
the stone meet at a height of 100m-80m-20m above the ground after 48.
Explanation:
Step 1: Given data
Initial velocity for upper stone, w=0. Initial velocity for lower stone, 259.
Height, h=100m.
Step 2: To find
When and where the two stones will meet.
Step 3: Calculate time and position of meeting
Let the stones meet at point A after time..
Displacement for upper stone:
Displacement for lower stone:
100-x-100-x-251-5² Adding displacement for upper stone and displacement for lower stone: 100 = 25 = 45 Substitute 4s for in displacement for upper stone: x=5² -54-80m Hence, the stone meet at a height of 100m-80m-20m above the ground after 48.
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