Question

solve this fast........... ​

Answer 1

EXPLANATION :

$\bull \: \: \: \: \: \LARGE \bf{Given :}$

$\to \sf{Initial \: Velocity \: = \: 28 \: m/s} \\ \\ \to \sf \: acceleration \: = - g \: m/ {s}^{2} \\ \\ \sf \: \to \: final \: velocity \: \: = 0 \: m/s$

$\sf \LARGE{let,}$

$\sf \: max \: hight \: reached \: = h \: meter \\ \sf \: and \: required \: time \: t \: sec$

$\bull \: \: \: \: \: \LARGE \bf{Required \: Formulas : :}$

$\sf \: \longrightarrow{v}^{2} = {u}^{2} + 2gx$

$\longrightarrow \: \sf \: x = ut + \frac{1}{2} g {t}^{2}$

$\longrightarrow \sf \: S_t = u + \frac{1}{2} g(2t - 1)$

$\bull \: \: \: \: \: \LARGE \bf{Solution :}$

$\sf \boxed {\sf{ {v}^{2} = {u}^{2} - 2gx}} \\ \\ \sf \: putting \: value \\ \\ \sf \implies {0}^{2} = {28}^{2} - 2 \times 9.8 \times h \\ \\ \implies \: \sf \: \underline{ \sf{h = 40}}$

$\sf \: Now, \: total \: time \: required \\ \\ \sf \: x = ut + \frac{1}{2} g {t}^{2} \\ \\ \implies \sf \: 40 = 28t - \frac{1}{2} \times 9.8 \times {t}^{2} \\ \\ \implies \: \underline{ \sf \: t = 1.18}$

$\sf \: before \: reaching \: max \: hight \\ \sf \: last \: sec \: \\ \boxed{\sf {S_t = u + \frac{1}{2} g(2t - 1)}} \\ \\ \sf \: putting \: value \\ \sf S_{0.18} = 28 - \frac{1}{2} \times 9.8(2 \times 0.18- 1) \\ \\ \implies \sf 31.136\: m$

$\sf \: velocity \: at \: S_{0.18} \\ \\ \boxed{ \sf{ {v}^{2} = {u}^{2} + 2gx}}$

$\sf \: velocity \: at \: S_{0.18} \\ \\ \boxed{ \sf{ {v}^{2} = {u}^{2} + 2gx}} \\ \\ \sf \implies {v}^{2} = {28}^{2} - 2 \times 9.8 \times 31.136 \\ \\ \therefore \boxed{ \boxed{{ \sf \: v = 13.18 \: m \: {s}^{ - 1} }}}$

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