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.

Asked by Vishal kumar | 1 year ago |  213

1 Answer

Solution :-

Given data:

(i) When the stone from the top of the tower is thrown,

Initial velocity u = 0

Distance travelled = x

Time taken = t

Therefore,

\( s = ut + \frac{1}{2}gt^2\)

x = 0 + (1/2)gt2

x = 5t ____________(a)

(ii) When the stone is thrown upwards,

Initial velocity u = 25 m/s

Distance travelled = (100 – x)

Time taken = t

\( s = ut + \frac{1}{2}gt^2\)

\((100-x) =25t + (1/2)\times10xt^2\)

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

From equations (a) and (b)

5t2 = 100 -25t + 5t2

t = (100/25) = 4sec.

After 4sec, two stones will meet

From (a)

x = 5t2 = 5 x 4 x 4 = 80m.

Putting the value of x in (100-x)

= (100-80) = 20m.

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

 

 

Answered by Shivani Kumari | 1 year ago

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