Chapter - 3 Gravitation

Q&A -Ask Doubts and Get Answers

Q
Gravitation

Question:

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:

Case(i): When the stone dropped from the top of the tower,

Initial velocity, u = 0

Let the distance travelled in time (t) = x

Using the second equation of motion,

S = ut + \frac{1}{2}gt^2

x = 0 \times t + \frac{1}{2}\times 10t^2

x = 5t^2 ---(1)

Case(ii): When the stone is thrown upwards,

Initial velocity u = 25 m/s

Distance travelled in time (t) = (100-x)

S = ut + \frac{1}{2}gt^2

100 - x = 25 \times t + \frac{1}{2}\times -10t^2

100 - x = 25t - 5t^2

x= 100 - 25t + 5t^2

x= 100 - 25t + 5t^2 ---(2)

From equations (1) and (2), we get

5t^2 = 100 - 25t + 5t^2

100 = 25t

t = 4\ sec

After 4sec, two stones will meet.

Now, by substituting the value of t in equation (1), we get

x = 5 \times (4)^2

x = 5 \times 16 = 80\ m

Putting the value of x in (100 - x), we get,

(100-80) = 20 m.

This means that after 4sec, 2 stones will meet at a distance of 20 m from the ground.


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