Question 17:
A stone is allowed to fall from the top of a tower 100 m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25 m/s. Calculate when and where the two stones will meet.
Answer:
Let the two stones meet after a time t.
(i) For the stone dropped from 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 m s−2
From the equation of motion,
(1)
(ii) For the stone thrown upwards:
Initial velocity, u = 25 m s−1
Let the displacement of the stone from the ground in time t be s’.
Acceleration due to gravity, g = −9.8 m s−2
Equation of motion,
(2)
The combined displacement of both the stones at the meeting point is equal to the height of the tower 100 m.
Therefore, the stones will meet after 4 s at a height (100 − 80) = 20 m from the ground
Latest Govt Job & Exam Updates: