Question

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Answer 1

v= 20 m /s

t=2 s

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Answer 2

$u =0 \: {ms}^{ - 1}$

$s = 20 \: m$

$a = 10 \: {ms}^{ - 2}$

Now,

${ \boxed{ \red{ {v}^{2} - {u}^{2} = 2as }}}\\ { \boxed{ \red{v = \sqrt{2as + {u}^{2} }}}} \\ \\ = > v = \sqrt{2(10)(20) + {0}^{2} } \\ = > v = \sqrt{2 \times 200 + 0} \\ = > v = \sqrt{400} = 20 \: {ms}^{ - 1}$

Again,

$t = \frac{v - u}{a} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: [°.° \: \:a = \frac{v - u}{t} ] \\ \\ \: \: = \frac{20 - 0}{10} \\ \\ \: \: \: = \frac{20}{10} \\ \: \: \: = 2 \: sec$

Hence,

${ \boxed{ \boxed{ \red{ \large{The \: \: ball \: \: will \: \: strike \: \: the \: \: ground }}}}}\\ { \boxed{ \boxed{ \red{ \large{with \: \: a \: \: velocity \: \: of \: \: 20 \: \: {ms}^{ - 1} \: \:in\: \: }}}}} \: \\ { \boxed{ \boxed{ \red{ \large{2 \: \: seconds}}}}}$