Question

Question #ffc23

Answer 1

`5/4s and 7.8125m`

Explanation:

Let the balls meet t sec after the second ball is thrown.
So the 1st ball will then complete (t+1) sec of its journey.

The imitial velocity of the 1st ball is zero .So it will fall a height `h=1/2*g*(t+1)^2`

The 2nd ball's initial velocity being `30m/s` it should fall the same height during t sec
So `h=30*t+1/2*g*t^2`

Inserting `g=10ms^-2` and equating two height we can write
`1/2*10*(t+1)^2=30t+1/2*10*t^2`
`=>5(t+1)^2-5t^2=30t`
`=>5(2t+1)=30t`
`=>2t+1=6t`
`=>t=1/4s`

So the ball will meet `(t+1)=1/4+1=5/4s` after the 1st ball starts.

During this time it will fall
`h=1/2*10*(5/4)^2=125/16m=7.8125m`