Question-58467

Question Number 58467 by azizullah last updated on 23/Apr/19

Commented by azizullah last updated on 23/Apr/19

$$\:\:\:\:\:\:\:\boldsymbol{\mathrm{please}}\:\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{with}}\:\boldsymbol{\mathrm{easy}}\:\boldsymbol{\mathrm{method}}. \\ $$

Commented by tanmay last updated on 23/Apr/19

$${what}\:{to}\:{find}…{question}\:{is}\:{not}\:{clear}… \\ $$

Commented by azizullah last updated on 23/Apr/19

dear sir make a quadratic equation and then solve?

$$\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{dear}}\:\boldsymbol{\mathrm{sir}}\:\boldsymbol{\mathrm{make}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{quadratic}}\:\boldsymbol{\mathrm{equation}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{then}}\:\boldsymbol{\mathrm{solve}}? \\ $$

Answered by tanmay last updated on 23/Apr/19

−3=14t−(1/2)×10×t^2 −3=14t−5t^2 5t^2 −14t−3=0 5t^2 −15t+t−3=0 5t(t−3)+1(t−3)=0 (t−3)(5t+1)=0 t=3second so the ball hit ground after 3 second

$$−\mathrm{3}=\mathrm{14}{t}−\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{10}×{t}^{\mathrm{2}} \\ $$$$−\mathrm{3}=\mathrm{14}{t}−\mathrm{5}{t}^{\mathrm{2}} \\ $$$$\mathrm{5}{t}^{\mathrm{2}} −\mathrm{14}{t}−\mathrm{3}=\mathrm{0} \\ $$$$\mathrm{5}{t}^{\mathrm{2}} −\mathrm{15}{t}+{t}−\mathrm{3}=\mathrm{0} \\ $$$$\mathrm{5}{t}\left({t}−\mathrm{3}\right)+\mathrm{1}\left({t}−\mathrm{3}\right)=\mathrm{0} \\ $$$$\left({t}−\mathrm{3}\right)\left(\mathrm{5}{t}+\mathrm{1}\right)=\mathrm{0} \\ $$$${t}=\mathrm{3}{second} \\ $$$${so}\:{the}\:{ball}\:{hit}\:{ground}\:{after}\:\mathrm{3}\:{second} \\ $$

Commented by azizullah last updated on 23/Apr/19

thank you sir answer is correct but solve by quadratic equation.

$$\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{thank}}\:\boldsymbol{\mathrm{you}}\:\boldsymbol{\mathrm{sir}}\:\boldsymbol{\mathrm{answer}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{correct}}\:\boldsymbol{\mathrm{but}}\:\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{by}}\:\boldsymbol{\mathrm{quadratic}}\:\boldsymbol{\mathrm{equation}}. \\ $$

Commented by MJS last updated on 23/Apr/19

but it is a quadratic! do you mean the formula? 5t^2 −14t−3=0 t=((−b±(√(b^2 −4ac)))/(2a))=((14±(√((−14)^2 −4×5×(−3))))/(2×5))= =((14±16)/(10))= { ((−(1/5))),(3) :}

$$\mathrm{but}\:\mathrm{it}\:{is}\:\mathrm{a}\:\mathrm{quadratic}! \\ $$$$\mathrm{do}\:\mathrm{you}\:\mathrm{mean}\:\mathrm{the}\:\mathrm{formula}? \\ $$$$\mathrm{5}{t}^{\mathrm{2}} −\mathrm{14}{t}−\mathrm{3}=\mathrm{0} \\ $$$${t}=\frac{−{b}\pm\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}{a}}=\frac{\mathrm{14}\pm\sqrt{\left(−\mathrm{14}\right)^{\mathrm{2}} −\mathrm{4}×\mathrm{5}×\left(−\mathrm{3}\right)}}{\mathrm{2}×\mathrm{5}}= \\ $$$$=\frac{\mathrm{14}\pm\mathrm{16}}{\mathrm{10}}=\begin{cases}{−\frac{\mathrm{1}}{\mathrm{5}}}\\{\mathrm{3}}\end{cases} \\ $$

Commented by azizullah last updated on 24/Apr/19

$$\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{thanks}}\:\boldsymbol{\mathrm{so}}\:\boldsymbol{\mathrm{much}}. \\ $$

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