YumYum247
  • YumYum247
elp me please!!!!!
Physics
jamiebookeater
  • jamiebookeater
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YumYum247
  • YumYum247
|dw:1441153687515:dw|
YumYum247
  • YumYum247
@IrishBoy123 Police help me!!! :)
YumYum247
  • YumYum247

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YumYum247
  • YumYum247
@dan815 Dan my love come rescue me from this terrible nightmare HUN!!!! :"D
YumYum247
  • YumYum247
Any idea?????????????????//
dan815
  • dan815
not sure, the question is a little too vague for me
dan815
  • dan815
does it mean launched by compression of spring or was it just thrown up from equilibriium position
dan815
  • dan815
the spring i take it is not attached to the ground
YumYum247
  • YumYum247
don't worry about the compression or anything, just work with the given variables...
dan815
  • dan815
im trying to determine what is given
YumYum247
  • YumYum247
the g = -9.81m/sec^2, Vi = 2.3m/sec, angle = 78deg
dan815
  • dan815
thats just a projectile problem it has nothing to with spring
YumYum247
  • YumYum247
but we have to solve the problem in projectile sense of view....i think.
dan815
  • dan815
oh theres nothing to that then just find V_y
dan815
  • dan815
Vy= sin78* 2.3 m/s a_y= -9.8m/s^2 h=? solve
dan815
  • dan815
|dw:1441158571844:dw|
YumYum247
  • YumYum247
and solve for what????
dan815
  • dan815
height, make sure u understand what is going on
dan815
  • dan815
it should be natural once u understand what is going on
dan815
  • dan815
you have an inital velocity, and ur acceleration is tellng u how much of that velocity u are losing per second
dan815
  • dan815
when the object is at its peak height, its veloicty is 0, so we want to see the time it has been travelling under that acceleration for
YumYum247
  • YumYum247
ok but i can't see the picture you just posted properly, some of the content is blocked/missing..... :(
dan815
  • dan815
once u get the time its been travelling, you can solve for the actualy distance it travelled in that time
dan815
  • dan815
zoom out to see everything
dan815
  • dan815
http://prntscr.com/8bltc5
YumYum247
  • YumYum247
ok...bit one more thing, there is a formula for this but i'm sure if that would work...let me post it...hold up :)
YumYum247
  • YumYum247
H=u2sin2(θ)2g
YumYum247
  • YumYum247
it looks something like this, le me draw.
YumYum247
  • YumYum247
|dw:1441159093595:dw|
YumYum247
  • YumYum247
is this right???????????
dan815
  • dan815
im thinking that is from equating the kinetic and potential energy together
dan815
  • dan815
see if both ways gives the same answer if it does then its right
YumYum247
  • YumYum247
but what if it doesn't???:O
dan815
  • dan815
1/2 m * V_Y^2 = mgh therefore V*sin ^2 (78)/2g = h
dan815
  • dan815
i think you are relying on formulas too much, you have to really understand physics or it wont work
YumYum247
  • YumYum247
ok what to do for now???? which one tho????
dan815
  • dan815
check both
YumYum247
  • YumYum247
aight.....!!! Thanks!!! :"D
dan815
  • dan815
i know whats gonna happen both ways, but i want you to explore
YumYum247
  • YumYum247
ok dear, i'll try to explore. Thank!!
YumYum247
  • YumYum247
what is the formula for projectile motion, i didn't get it for a sec.... t = Vf +Vi + 1/2(G+t^2)
YumYum247
  • YumYum247
dan815
  • dan815
|dw:1441161107807:dw|
dan815
  • dan815
|dw:1441161320080:dw|
dan815
  • dan815
u can do this integral too to see if gives u the same distance
dan815
  • dan815
http://www.wolframalpha.com/input/?i=t%3D2.3sin%2878%29%2F%289.8%29%2Ch%3D2.3sin%2878%29+*+t%2B1%2F2%28-9.8%29*t^2
YumYum247
  • YumYum247
nooooooooooooooooooooooooooooooooooooooi 2 is not the power.....
dan815
  • dan815
http://www.wolframalpha.com/input/?i=h%3D2.3^2*sin^2%2878%29%2F%282*9.8%29
YumYum247
  • YumYum247
2 multiplies with the sign...i have the formula sheet with me rn.
dan815
  • dan815
both ways same answer
dan815
  • dan815
the formula is wrong
dan815
  • dan815
i showed u how the formula is derived
dan815
  • dan815
it comes from kinetic1+potenrial1= kinetic2+ potential2
YumYum247
  • YumYum247
ok......
dan815
  • dan815
so in the beginning theres only kinetic as the height is 0 and in the end theres only potential as the velocity in y direction is 0
dan815
  • dan815
1/2mv^2=mgh m cancel both sides 1/2 V^2=gh so h=1/2v^2/g
YumYum247
  • YumYum247
yah makes sense......
dan815
  • dan815
or h=v^2/2g
dan815
  • dan815
now v^2 is the V_y^2so
YumYum247
  • YumYum247
wait what do you mean by Vy tho...vertical velocity O_o
dan815
  • dan815
V_y^2=(Vi*sin 78)^2 =Vi^2 * sin^2 (78)
dan815
  • dan815
yes Vy is the vertical velocity
dan815
  • dan815
Vi is more like the speed or the veloicty in that specific direction
YumYum247
  • YumYum247
ok thanks a lot.....i mean alot. :)
dan815
  • dan815
okay but u better understand this stuff
dan815
  • dan815
the most important thing is you are able to visualize and work through it in your head
dan815
  • dan815
or paper
YumYum247
  • YumYum247
yah i'm doing it....it takes time bro, gimme sumtime bro.......
YumYum247
  • YumYum247
|dw:1441162910616:dw|
YumYum247
  • YumYum247
@dan815 i'm stuck!!!! :"(
dan815
  • dan815
what kinda equation is that
dan815
  • dan815
its h(t) = 0 + v_y*t+1/2 * a*t^2
dan815
  • dan815
h(t) is notation for h as a function of t
dan815
  • dan815
Vy= vi*sin (78) t=Vy/9.8 h = 0 + v_y*t+1/2 * a*t^2
YumYum247
  • YumYum247
i didn't post the whole thing but i followed your steps.....
YumYum247
  • YumYum247
Vy= vi*sin (78) t=Vy/9.8 h = 0 + v_y*t+1/2 * a*t^2 if you simplify this you get something i posted.....
dan815
  • dan815
\[V_y= vi*sin (78)\\ t=V_y/9.8 \\ h= 0 + v_y*t+1/2 * a*t^2\]
YumYum247
  • YumYum247
that's a quadratics expression.
YumYum247
  • YumYum247
if you rearrange it.
dan815
  • dan815
its not a quadratic u know what t is already u plug that in
YumYum247
  • YumYum247
omg...i didn't see that....sory....:) let me try again....
dan815
  • dan815
Vi=2.3 Vy=2.3*sin(78) t=Vy/9.8 h=0+Vy*t + 1/2 a*t^2
YumYum247
  • YumYum247
don't say anything just let me take a look at this with my eyes and brain open...
YumYum247
  • YumYum247
i got it....
YumYum247
  • YumYum247
Thanks!!!
dan815
  • dan815
check with h=Vy^2/(2g)
YumYum247
  • YumYum247
yah that works for me....
YumYum247
  • YumYum247
h = vertical velocity/2g gives me the right answer....
YumYum247
  • YumYum247
Anyways don't stress yourself..... go watch a porn to cool off a lil.... :"D i'll get this done on my own now!! :)
YumYum247
  • YumYum247
btw the example you gave me on wolframe didn't have the time input for T.
YumYum247
  • YumYum247
|dw:1441164804108:dw|
YumYum247
  • YumYum247
@dan815 please take a look at this.
YumYum247
  • YumYum247
i have to use quadratics formula here....
dan815
  • dan815
http://www.wolframalpha.com/input/?i=t%3D2.3sin%2878%29%2F%289.8%29%2Ch%3D2.3sin%2878%29+*+t%2B1%2F2%28-9.8%29*t^2
YumYum247
  • YumYum247
yes i get that...but what happened at the very last step....
YumYum247
  • YumYum247
idk if the website used quadratics formula, or what to get the answe!!!
YumYum247
  • YumYum247
im stuck at the very last part....
YumYum247
  • YumYum247
|dw:1441165539159:dw|
YumYum247
  • YumYum247
never mind i got it....
YumYum247
  • YumYum247
Thanks !!

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