Question #79341

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Question #79341

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Answer 1 · 2015-02-13T01:18:19

$V_{1} = 26.34 m / s$ $V_1=26.34m//s$

$V_{2} = 29.2 m / s$ $V_2=29.2m//s$

I have assumed that by $V_{1}$ $V_1$ and $V_{2}$ $V_2$ you mean the final velocity of the ball in each case.

We can use this equation of motion:

$v^{2} = u^{2} + 2 a s$ $v^(2)=u^(2)+2as$

$v =$ $v=$ final velocity

$u =$ $u=$ initial velocity

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

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

In the first case:

$V_{1}^{2} = 0 + (2 \times 9.8 \times 35.4) = 693.84$ $V_1^(2)=0+(2xx9.8xx35.4)=693.84$

$V_{1} = \sqrt{693.84} = 26.34 m / s$ $V_1=sqrt(693.84)=26.34m//s$

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

So:

$V_{2}^{2} = {(12.6)}^{2} + (2 \times 9.8 \times 35.4) = 852.76$ $V_2^(2)=(12.6)^(2)+(2xx9.8xx35.4)=852.76$

$V_{2} = \sqrt{852.76}$ $V_2=sqrt852.76$

$V_{2} = 29.2 m / s$ $V_2=29.2m//s$

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Question #79341

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