+14915 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
\(t= {-b \pm \sqrt{b^2-4ac} \over 2a}\)
\(\large t = {90.4812 \pm \sqrt{8186.8472-4\times 4.905\times (-100)} \over 2\times 4.905}\)
\(\large t = {90.4812 \pm \sqrt{10148.8472} \over 9.81}\)
\(\large t=\frac{90.4812\pm 100.7415}{9.81}\) \((-100.7415\ entf\ddot allt)\)
\(t= 19.493\)
!
+14915 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
\(t= {-b \pm \sqrt{b^2-4ac} \over 2a}\)
\(\large t = {90.4812 \pm \sqrt{8186.8472-4\times 4.905\times (-100)} \over 2\times 4.905}\)
\(\large t = {90.4812 \pm \sqrt{10148.8472} \over 9.81}\)
\(\large t=\frac{90.4812\pm 100.7415}{9.81}\) \((-100.7415\ entf\ddot allt)\)
\(t= 19.493\)
!
