Recent content by red_viper_88

thank you so much for all your time, help, and patience![/color]

so which is the correct way to do it? and what would u advise me to put as my final answer?

yup! u r right! so the answer of 219 secs matches urs? or is it 220 secs? or with the formula u told me to use, would be: t = (650 x 10^3) / 3157 = 206 secs can u please double check that? thanks!

i had a friend help me finish it up, can u please double check the answer! thanks! vertical component = v sin 20 3km sin 20 s = ut + 1/2at^2 s = 0 u = 3 a = -9.8 thus, 4.9t^2 - 1076t = 0 t(4.9t - 1076) = 0 4.9t = 1076 t = 1076 / 4.9 =...

so no way to help get me started?

is it the correct number?

oh wow! i really like the formula u gave me better! let me try that! so 4.9 (650 x 10^3 / 0.94v) = 0.34v then 4.9 (650 x 10^3) = 0.34v x 0.94 v then 4.9 (650 x 10^3) = 0.34 x 0.94 v^2 then v^2 = 4.9 x (650 x 10^3) / 0.34 x 0.94 then v = 3156.83? that is one big number!

i got 1/2g = v sin(20) / t...is that right? thus i would get 4.9 = v sin 20 / t and then i would substitute that equation i made for t? and get 4.9 on one side and 0.34v / (650 x 10^3 / 0.94v) on the other side?

i can't seem to get v alone! = -4.9 (650 x 10^5 / 0.88 v^2) + 0.34 v (650 x 10^3 / 0.34 v) i simplified the second equation a lil more, so i can work with it easier before substituting

so for the first one t = (650 * 10^3) / v cos (20) ?

i don't understand how i combine them

how is it solvable with 2 unknowns?

displacement = uniform velocity x time this uniform velocity = displacement / time constant velocity = 650 x 10^3 m / time ?? i'm very lost! there is another equation i found is v = at...but I'm still messed up with the t variable! i can't substitute anything for that in order to find...

gravity? so I'm substituting -9.8 for acceleration (g)? what is s in the equations u listed? displacement? and i don't know what time is...so I'm kind of confused as to how to use those equations

I believe there is an acceleration in the y-direction, i don't know what it is and how to find it. And acceleration in this case would be constant. I don't think there is an acceleration in the x-direction. Just a velocity. I do not know the "kinematic equations for constant acceleration and...