20 = v*t*sin35 - 4.9*t^2
v*t*cos35 = 130
2007-09-04 14:17:36 · 4 answers · asked by J 1 in Science & Mathematics ➔ Mathematics
20 = vt sin 35 - 4.9*t^2
vt sin 35 = 20 + 4.9*t^2 - 1st equation
vt cos35 = 130 - 2nd equation
take 1st eqn divided by 2nd eqn, the vt will cancel out, sin/cos = tan so...
tan 35 = ( 20 + 4.9*t^2) / 130
130 tan 35 = 20 + 4.9*t^2
(130 tan 35 - 20)/4.9 = t^2
t = sqrt(14.49)
= +- 3.807 (4s.f.)
substitute into second equation
v(3.807)cos 35 = 130
v = 41.7(3.s.f)
or
v(-3.807) cos 35 = 130
v = -41.7(3s.f.)
2007-09-04 14:28:24 · answer #1 · answered by Charliemoo 5 · 0⤊ 0⤋
1) t = 130/v*cos35
2) v*tsin35 -4.9t^2 = 20
1) => 2) v* 130/v*cos35*sin35-4.9(130/v*cos35)^2 = 20
Just solved this quadratic equation for v
with this formula : v = -b ±sqrt(b^2-4ac)/2a
standard form : av^2 + bv + c = 0
2007-09-04 21:33:35 · answer #2 · answered by frank 7 · 0⤊ 0⤋
use substitution from the second eq
t = 130 /( v * cos 35 )
20 = v * 130 * sin 35 / (v * cos35) - 4.9 * (130 / (v * cos 35)) ^ 2
Use a calculator, you can eliminate most of the const values.
2007-09-04 21:27:59 · answer #3 · answered by norman 7 · 0⤊ 0⤋
yo
2007-09-04 21:23:05 · answer #4 · answered by Ashwin 2 · 0⤊ 0⤋