Question #79341

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Answer 1 · 2015-02-13T01:18:19

$V_1=26.34m//s$

$V_2=29.2m//s$

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:

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

$V_2=sqrt852.76$

$V_2=29.2m//s$