Question

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

Answer 1

HEYA !!!

## HERE'S THE SOLUTION -

Height of the tower = 100m

g = 10 m/s 2

velocity of projection of ground projectile = 25 m/s

Velocity of projection of the tower projectile = 0

Let the particles meet at time t and at height h

Thus,

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.

I hope You found My Answer Helpful

:-)