Ok so I did:
dy=v1yt+1/2ayt2
-1.1=v1yt-4.905t2 9.81 for g here.
4.905t2-v1yt-1.1
t=-v1y+/-(((-v1y)2-4(4.905)(-1.1))^.5)/2(4.905)
t=-vsinθ+(((-vsinθ)2+21528)^.5)/9.81 (has to be positive or time will be negative)
then from before:
1/2mv2=1/2kx2
v=((kx2)/m)^1/2 but I only want the x speed...
Sorry about the t.
if s0 stands for displacement (it should be -1.1 then?), then I set s = 0 and do the quadratic?
EDIT: Also why is yours +1/2at^2 where as mine and cyosis is -1/2at^2 ?
Yes it is an experiment I am doing at home and I am not too worried about resistance.
Ok, I am going to assume it is
v2=v1+at because I need time, and I don't have distance.
t=v1sinθ/a because v2 is 0 halfway up
so
t=2(vsinθ/a) still am not accounting for the 1.1m off the ground (not sure how...
Homework Statement
So I made a spring launcher that will fire marbles. I need to derive a general formula for finding range for any given angle and x value.Homework Equations
1) v22=v12+2ad
2) Ek=Es
3) 1/2mv2=1/2kx2
4) d=v*t
This projectile was shot from 1.1m off of the ground.
The Attempt at...
Homework Statement
Find the max height of a projectile fired at an angle of 30 degrees if it's inital speed is 0.714m/s.
(this is a lab so I have other data that might be needed)
Homework Equations
V22=V12+2ad
The Attempt at a Solution
V22=V12+2ad
0=(0.714sin30o)2+2(-9.81)(d)
d=19.62-0.127449...