Question

plz solve this ........

Answer 1

$d = u \sqrt{ \frac{2h}{g} } \\ 5 =u \sqrt{ \frac{8}{10} } \\ 25 = {u}^{2} \times \frac{8}{10} \\ u = \sqrt{ \frac{250}{8} } \\ u = \frac{5 \sqrt{5} }{2}$
Hope this helps you

Answer 2

❤️Hope u like my process❤️
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Formula to be used:-
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$\bf \: = > range(x) = v \sqrt{ \frac{2h}{g} } \\ \\ \bf = > \: time \: \: period(t) = \sqrt{ \frac{2h}{g} } \\ \\ \bf \: (where \: \: v = velocity. \\ \\ \bf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: h = height \\ \\ \bf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: g = gravity = 9.8 \: m \: {s}^{ - 2} )$

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Given:-
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=> Range (x) = $\bf \: 5 \: \: m$

=> Height (h) = $\bf \: 4 \: \: m$
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So,
$\bf = > x = v \sqrt{ \frac{2h}{g} } \\ \\ or. \: \: v = x \sqrt{ \frac{g}{2h} } = 5 \times \sqrt{ \frac{9.8}{2 \times 4} } \\ \\ or. \: \: \bf v = \frac{5 \times 7}{2 \sqrt{10} } = 5.53 \: m \: {s}^{ - 1}$

$\bf = > t = \sqrt{ \frac{2h}{g} } = \sqrt{ \frac{2 \times 4}{9.8} } = 0.9 \: s$
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❤️Hope this is ur required answer❤️

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