A ball is dropped from the jumping board of a swimming pool, which is at a height of 20 m. Second ball is thrown from the same board after one second with initial velocity u. Both the balls hit the water together. What was the initial velocity with which the second ball was thrown? Do they hit water with the same velocity? Explain your answer.

Explanation

For the first ball:
u=0
`g=-10m//s^(2)`
S=-20m
`v=v_(1)=?`
`t=t_(1)=?`
Using `S=ut+1/2"gt"^(2)` we get
`-20=0xxt+1/2(-10)t^(2)`
or `-20=-5t^(2)`
or `t^(2)=4`
or `t=sqrt(4)=2s`
Also `v=u+"gt"`
Also `v=u+"gt"`
`:.v_(1)=0+(-10)xx2`..........1
or `v_(1)=-20m//s`
For the second ball
`u=?`
`g=-10m//s^(2)`
`S=-20m`
`g=(2-1)=1s`
( `:.` both the ball hit the water together)
Using `S=ut+1/2"gt"^(2)` we get
`-20=uxx1+1/2(-10)(1)^(2)`
or `-20=u-5`
or `u=-20+5=-15m//s`
`:.` the second ball was thrown with a velocity of 15 m/s
Using `v=u+"gt"` we have
`v_(2)=-15+(-10)xx1`
or `v_(2)=-25m//s` .....2
comparing 1 and 2 we find that the two balls will not hit water with the same velocity.