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-100=150sin(37.1)t-1/2(9.81)t^2

 Sep 25, 2017

Best Answer 

 #1
avatar+8049 
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physics problem help. find t

-100=150sin(37.1)t-1/2(9.81)t^2

 

Hello Guest!

 

\({\color{BrickRed}-100=150\cdot sin(37.1°)\cdot t-\frac{1}{2}\cdot 9.81\cdot t^2}\\ \frac{1}{2}\cdot 9.81\cdot t^2 -150\cdot sin(37.1°)\cdot t-100=0\\ 4.905t^2-90.4812t-100=0\)

   a                   b                c

\(t= {-b \pm \sqrt{b^2-4ac} \over 2a}\)

 

\(\large t = {90.4812 \pm \sqrt{8186.8472-4\times 4.905\times (-100)} \over 2\times 4.905}\)

 

\(\large t = {90.4812 \pm \sqrt{10148.8472} \over 9.81}\)

 

\(\large t=\frac{90.4812\pm 100.7415}{9.81}\)    \((-100.7415\ entf\ddot allt)\)

 

\(t= 19.493\)

 

laugh  !

 Sep 25, 2017
 #1
avatar+8049 
+2
Best Answer

physics problem help. find t

-100=150sin(37.1)t-1/2(9.81)t^2

 

Hello Guest!

 

\({\color{BrickRed}-100=150\cdot sin(37.1°)\cdot t-\frac{1}{2}\cdot 9.81\cdot t^2}\\ \frac{1}{2}\cdot 9.81\cdot t^2 -150\cdot sin(37.1°)\cdot t-100=0\\ 4.905t^2-90.4812t-100=0\)

   a                   b                c

\(t= {-b \pm \sqrt{b^2-4ac} \over 2a}\)

 

\(\large t = {90.4812 \pm \sqrt{8186.8472-4\times 4.905\times (-100)} \over 2\times 4.905}\)

 

\(\large t = {90.4812 \pm \sqrt{10148.8472} \over 9.81}\)

 

\(\large t=\frac{90.4812\pm 100.7415}{9.81}\)    \((-100.7415\ entf\ddot allt)\)

 

\(t= 19.493\)

 

laugh  !

asinus Sep 25, 2017

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