Question

From the top of a tower. 200m in height. A ball is dropped at the same time another ball is projected vertically upwards from the ground with a velocity of 50m/sec. Find when and where will the two balls 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 = 50 × t - 1/2 × 10 × t^2

or, s = 50 t - 5 t^2

As initially the gap between both the stone is 200 m ... so they both will together travel a distance of 200 m to meet

so, ( 5 t^2 ) + ( 50 t - 5 t^2 ) = 200

or, 50 t = 200

or, t = 200/50 = 4 sec

so, they will meet after 4 sec.

S for falling stone = 5 × 4^2 = 80 m from the top of the tower.

s for projected stone = 50×4 - 5×4^2

= 200 - 80 = 220 m from the ground.

Answer 2

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