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 25m/s. calculate when and where the two stones will meet.
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Answer 1

Let they meet after t seconds

When stone falls from top

let they meet at a distance h from top

h= 1/2 g t^2

Now when another stone is projected upwards with velocity = 25m/s

Its displacement when they meet = 100 - h

So

100 - h= 25t - 1/2 g t^2

put h= 1/2 gt^2

100 - 1/2 gt^2 = 25t - 1/2 g t^2

100= 25 t

t= 4 sec

After 4 sec they meet

and they meet at

h= 1/2 g t^2 = 5( 4)^2 = 80 m

So they meet at 80 m from top

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) t²

or, S = 5t²

◆ ii) For stone projected upward

(100 - S) = 25t + 1/2 (-10) t²

= 25t - 5t²

Adding i) and ii) , We get

100 = 25t

or t = 4 s

Therefore, Two stones will meet after 4 s.

◆ ¡¡¡) Put value of t = 4 s in Equation i) , we get

S = 5 × 16

= 80 m.

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