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