Question

A stone is allowed to fall from the top of a tower 100m high and at the same time together stone is projected vertically upwards from the ground with a velocity of 25m/s . Calculate when and where the two stones meet.​

Answer 1

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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.

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Answer 2

Answer:-

U1 = 0

let the stones meet at a height of h about the ground stone 1 covers a distance (100-h) ; g = 100ms²

(100-h) = 0 + 1/2 gt²

=> 100-h = 5t² ---------1.

• Therefore V2 = 25 m/s.

》stone 2 covers a distance

g= -10m/s.

h= U2t - t/2 gt²

=> h = 25t - 5t² -----------2.

from (i) & (ii) , we get

100 - 25t + 5t² = 5t²

or t = 4s ( stones meets 45 after they are thrown h = 25 (t) - 5(t)² = 100-80 =20m

above the ground.