Finding time for air drag projectile at 45 degrees

[Matt Jacques]

Since my other thread is lacking attention, perhaps it is more suitable here:

I inserted some fixed constants and multiplied out

(48/.5)(1-^e-.5t)(sin 45) + (9.8/.25)(1-(.5t) - e^-.5t) = 0

(96 - 96e^-.5t)(sin 45) + 39.2(1-(.5t)-e^-.5t) = 0

67.88 - 67.88^e-.5t + 39.2 - 19.6t - 19.6e^-.5 = 0

107.2 - 87.48e^-.5t - 19.6t = 0

107.2 - 19.6t = 87.48^e-.5t

log(107.2 - 19.6t) = log(87.48^e-.5t)

log107.2 - log19.6t = -.5tLog(87.48)

2.030194 - log19.6t = -.5t(1.94198)

1.045463 - log19.6t = -.5t

-(1.045463 - log19.6t = -.5t)

-1.045463 + log19.6t = .5t

log19.6t = .5t + 1.045463

10^(.5t + 1.045463) = 19.6t

This is where I am stuck.

 

[quantumdude]

Matt,

Could you post the original problem, or a link to your other thread? I don't see what you're trying to do with this.

 

[Matt Jacques]

https://www.physicsforums.com/showthread.php?s=&threadid=1100

I just need the formula to find the time of a projectile influenced by air drag launched from either y=0 or some height