Help urgent please, some physics some math
@Abhishek619 or anyone
@Loser66 thanks for having a look i have no clue where to start yet
May be I cannot help. Anyway, I am trying
ty i have no hope i can do it myself alone :(
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y=h+s(cotb) use equations of motion. y=(Vo sin (alpha) (t)+(1/2)(-g)t^2 and also, (Vo cos (alpha)) (t)=d+s get s from there.
so, you can get the time the stone travels.
im trying to digest the equation both..
agree with Abhishek619
i get Loser66's vertical component equations
to me, after getting the time t, apply to distance equation from horizontal line. It says x = x_0 + V_0x t + 1/2 at^2 , where x_0=0, a =0 and X = d+s
and V_0x = V_0 cos 49
how to get time, i didnt get Abhishek's y = h+scot(b) equation sorry :(
hehehe, I go backward.
vertical : y = 20sin(49) - 5 t^2
horizontal : x = 20cos(49)*t
wher to go from here
what is x? x is the horizontal distance which the particle has covered. what is that? it is the sum of h and s. do you agree?
^^ yes... x = h+s when the stone hits the incline
sorry, x = d+s
oh sorry! my bad! good going. t=d+s/ (Vo cos (alpha)) substitute t in y=(Vo sin (alpha) (t)+(1/2)(-g)t^2 you get a quadratic equation in terms of 's'. now solve that quadratic equation and get the value of s.
oww it looks easy, ima try
t = (15+s)/(20 cos(49)) y = 20sin(49) - 5 t^2
but 2 equaitons with 3 unknowns : t, y, s ?
y=h+ s(tan (beta))
ugh im real dumb
t = (15+s)/(20 cos(49)) y = 20sin(49) - 5 t^2 y= 6+ s(tan (30))
3 equations , 3 unknowns anbody can solve lol you're awesome thanks a lot !!!!!
lol yea. you're welcome!!
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