Question

A stone is allowed to fall from the top of a tower 100m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25ms-1.calcuate when and where the two stones meet

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.