- #1
Tschew
- 11
- 0
Hi!
I'm slightly stuck with an approximation of tmax, the time at which a ballistic projectile reaches maximum height. (projectile shot straigh upwards, no x motion)
So, by integrating the motion equation
m * dv/dt = -mg -bv
(where b is an air drag constant) I've created:
v(t) = (u + mg/b)*e^(-bt/m) - mg/b
which at tmax (t at maximum height) must be zero, which then gives:
tmax = (-m/b) ln [ (mg/b) / (mg/b + u) ]
where u is the initial velocity in y
Now when m >> b*tmax, tmax = u/g , howver, I can't get that!
all I get is (using e^x = 1 + x + x^2/2! +...):
tmax = u / (mgb/bm + ub/m)
thanks for any help!
-tschew
I'm slightly stuck with an approximation of tmax, the time at which a ballistic projectile reaches maximum height. (projectile shot straigh upwards, no x motion)
So, by integrating the motion equation
m * dv/dt = -mg -bv
(where b is an air drag constant) I've created:
v(t) = (u + mg/b)*e^(-bt/m) - mg/b
which at tmax (t at maximum height) must be zero, which then gives:
tmax = (-m/b) ln [ (mg/b) / (mg/b + u) ]
where u is the initial velocity in y
Now when m >> b*tmax, tmax = u/g , howver, I can't get that!
all I get is (using e^x = 1 + x + x^2/2! +...):
tmax = u / (mgb/bm + ub/m)
thanks for any help!
-tschew