Question

A ball is dropped from a bridge of 122.5 m above a river .Afterthe ball has been falling for two seconds ,a second ball is thrown straight down after it . Initial velocity of the second ball so that body hit the water at the same time is

Answer 1

For ball 1

v²=u²+2gh

v=√2gh

v=√2×10×122.5

v=√2500

v=50 m/s

v=gt

50=10t

t=5 sec.

For ball 2

Ball 2 is released after two sec.

∴ Time of flight= 3 sec.

s=ut+1/2×at²

122.5=3u +45

3u=75

u= 25 m/s

Answer 2

s =ut+1/2 gt^2

122.5=1/2×9.8×t^2

122.5=4.9×t^2

122.5/4.9=t^2

t^2=25

t=5

next ball cover 3 sec to hit water at same time

s=ut+1/2t^2

122.5=3u+4.9×3^2

122.5=3u+4.9×9

122.5=3u+44.1

122.5-44.1=3u

78.4=3u

u=26.133 m/s

u = 26.1m/s