Question

Question #79341

Answer 1

`V_1=26.34m//s`

`V_2=29.2m//s`

I have assumed that by `V_1` and `V_2` you mean the final velocity of the ball in each case.

We can use this equation of motion:

`v^(2)=u^(2)+2as`

`v=` final velocity

`u=` initial velocity

`a=` acceleration which in this case I will assume to be equal to the acceleration due to gravity which is given by `g=9.8m//s//s`.

`s=` displacement (how far the ball drops).

In the first case:

`V_1^(2)=0+(2xx9.8xx35.4)=693.84`

`V_1=sqrt(693.84)=26.34m//s`

In the second case the initial velocity `u=12.6m//s`

So:

`V_2^(2)=(12.6)^(2)+(2xx9.8xx35.4)=852.76`

`V_2=sqrt852.76`

`V_2=29.2m//s`