Question

How do I fix this equation?

r= v^2/(g)tan( )

Answer 1

well, you need an angle. How you fix it depends on what you want it to do.

If it has anything to do with a trajectory, you might want to read the excellent article in wikipedia.

Answer 2

I am not sure what you mean.
Perhaps it has something to do with how high a projectile goes
kinetic energy due to vertical velocity component at ground = potential energy at top
(1/2) m v^2 = m g h
so
h = v^2/ (2g)
now if you launched at angle T from the ground at speed s
then v = s sin T
and u = horizontal component of speed = s cos T
now how long did it take to reach that height h?
average speed up = v/2
so time going up = 2 h /v
the total time in the air including the fall is then 4 h/v = 4 v^2/(2gv) = 2v/g
so how far did it go along the ground, range r?
time times horizontal component of speed = Time in air * s cosT
r= 2 v/g * s cos T
but v = s sin T
r= (2/g)s^2 sin T cos T = (s^2/g) sin 2T

Answer 3

I derived this equation from Banking of Curves without friction which is:

mgtan( )=mv^2/r

The problem was asking for the radius which r, so I came up with the formula:

r=v^2/gtan()

I do not know how the rules of inverse apply to the derivation of the formula.

Answer 4

All other variables are given which are:

v= 15 m/s
g(ravity)=9.8 m/s^2
Angle= 20 degrees (tan(20))

Answer 5

oh
call force normal to track = F
horizontal force = m v^2/R = F sin 20
vertical force = m g = F cos 20
so
F= m v^2 /R sin 20 = m g/cos 20
g R sin 20 = v^2 cos 20
R = (v^2/g ) cos 20/sin 20 = v^2/ (g tan 20)