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Let the two stones meet after a time t.
(i) For the stone dropped from the top of the tower:
Initial velocity, u = 0
Let the displacement of the stone in time t from the top of the tower be s.
Acceleration due to gravity, g = 9.8 ms-2
From the equation of motion,
(ii) For the stone thrown upwards:
Initial velocity, u = 25 ms-1
Let the displacement of the stone from the ground in time t be s'.
Acceleration due to gravity, g = -9.8 ms-2
From the equation of motion,
The combined displacement (s + s') of both the stones at the meeting point is equal to the height of the tower 100 m.
From eqs. (1) and (2), we get,
s + s' = 4.9 t2 + 25 t - 4.9 t2
100 = 25 t
In 4 s, the falling stone has covered a distance given by equation (1) as
s = 4.9 t2 = 4.9 × (4)2 = 78.4 m
Therefore, the stones will meet after 4 s at a height (100 - 78.4) = 21.6 m from the ground. hope u like it ,
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(i) For the stone dropped from the top of the tower:
Initial velocity, u = 0
Let the displacement of the stone in time t from the top of the tower be s.
Acceleration due to gravity, g = 9.8 ms-2
From the equation of motion,
(ii) For the stone thrown upwards:
Initial velocity, u = 25 ms-1
Let the displacement of the stone from the ground in time t be s'.
Acceleration due to gravity, g = -9.8 ms-2
From the equation of motion,
The combined displacement (s + s') of both the stones at the meeting point is equal to the height of the tower 100 m.
From eqs. (1) and (2), we get,
s + s' = 4.9 t2 + 25 t - 4.9 t2
100 = 25 t
In 4 s, the falling stone has covered a distance given by equation (1) as
s = 4.9 t2 = 4.9 × (4)2 = 78.4 m
Therefore, the stones will meet after 4 s at a height (100 - 78.4) = 21.6 m from the ground. hope u like it ,
Mark brainliest plz
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