Question

a stone is allowed to fall from the top of a tower 300m height and at the same time another stone Is projected vertically up from the ground with a velocity 100ms.find when and where the two stones meet?

Answer 1

let after time t the both stone will meet.

distance travelled by the stone falling from the tower in time t

s = u t + 1/2 g t^2

or, s = 0 + 1/2 × 10 × t^2 = 5 t^2

Also, distance travelled by the stone thrown up from ground in time t =

s = u t + 1/2 (-g) t^2

or, s = 100 × t - 1/2 × 10 × t^2

or, s = 100 t - 5 t^2

As initially the gap between both the stone is 300 m ... so they both will together travel a distance of 300 m to meet

so, ( 5 t^2 ) + ( 100 t - 5 t^2 ) = 300

or, 100 t = 300

or, t = 300/100 = 3 sec

so, they will meet after 3 sec.

S for falling stone = 5 × 3^2 = 45 m from the top of the tower.

s for projected stone = 100×3 - 5×3^2

= 300 -45 = 255 m from ground.

Answer 2

Hope it helps you......