Question

A ball dropped from bridge of 122.5 m above the river. after the ball has been failing for.two secounds , a secound ball is thrown straight down after it . initial velocity of secound ball so that both hit the water at same.time is

Answer 1

let velocity be x ,accleration is 9.8 and distance is 122.5 the second ball is thrown after 2 seconds so the distance covered by the first ball in this time would be s= u×t+ 0.5 ×a×t×t applying this we get s as 19.6 it means that the second ball has to cover 19.6 m more to fall at same time so let ( assumption) the total distance travelled by the second ball be 122.5+ 19.6= 142.1 m so now assume that the first and second ball were thrown at same time but different heights which is 122.5 and142.1 m so using third law of motion which is v^-u^= 2 ×a × s we get the velocity as 52.77 m\ s which is the answer

Answer 2

Answer:26.133 seconds

Explanation:

Time taken by first ball to hit water can be found as

s= ut + ½ gt2

122.5 = ½ * 9.8 * t2

t2= 25

t =5

Now the next ball must cover this distance in 3 s to hit water at same time.

122.5 = 3 u + ½ * 9.8 * 32

122.5 = 3u + 44.1

3u = 78.4

u= 26.133 m/s

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