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 1

let "t" = time after which both stones meet

"S" = distance travelled by the stone dropped from the top of tower

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(100-S) = distance travelled by the projected

stone.

i) For stone dropped from the top of tower

-S = 0 + 1/2 (-10)t?

or, S = 5t2

• ii) For stone projected upward

(100 - S) = 25t + 1/2 (-10) t? =

= 25t - 5t2

Adding i) and ii) , We get

100 = 25t

or t = 4 s

Therefore, Two stones will meet after 4 s.

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iii) Put value of t = 4 s in Equation i), we

get

S = 5 5 x 16

= 80 m.

1.1 47

Thus, both the stone will meet at a distance of 80 m from the top of tower.

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Answer 2

let "t" = time after which both stones meet

"S" = distance travelled by the stone dropped from the top of tower

__________________________________________

(100-S) = distance travelled by the projected

stone.

i) For stone dropped from the top of tower

-S = 0 + 1/2 (-10)t?

or, S = 5t2

• ii) For stone projected upward

(100 - S) = 25t + 1/2 (-10) t? =

= 25t - 5t2

Adding i) and ii) , We get

100 = 25t

or t = 4 s

Therefore, Two stones will meet after 4 s.

________________________________________

iii) Put value of t = 4 s in Equation i), we

get

S = 5 5 x 16

= 80 m.

1.1 47

Thus, both the stone will meet at a distance of 80 m from the top of tower.

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