Question

A tower of height 100m.a stone is thrown up from the foot of water with a velocity of 20m/s and at the same time another stone is dropped from the too of the tower the two stones will meet after

Answer 1

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.
Thank you.