Air resistance of a thrown object

[devanlevin]

an object thrown vertically upwards into the air takes shorter time to reach its peak height than it does to return to the height of the throw, is this true, how can it be, on the way up it has both the airs resistance ang gravity working against it whereas on the way down only the resistance,

is the speed of the object necessarily the same when it passes the beginning point on its descent as it was when it was thrown(Vo=Vf)? even in a case where air resistance is taken into account? if so i think i understand, - because the acceleration is lesser on the descent(because the forces are working in opposite directions) the time is greater. is this correct?
how can i summarise this using the equations of work and energy?

 

[PhanthomJay]

When you have air resistance forces acting, Vo at the release point and Vf when it returns to the release point cannot be the same. For one thing, you could throw a projectile up at some very high speed, but when it returns, it can't be traveling faster than its terminal velocity. Otherwise, i think you're on the right line of reasoning. Consider that when the projectile is thrown up, as you note, both gravity and air drag forces oppose the motion, causing it to de-celerate at a rate greater than g. This rapidly decreases the projectiles speed until it comes to a stop. It's actual deceleration will vary from some number greater than g, down to g, (at its peak), during the rise, due to the complex nature of the non constant drag force. When it now stops and starts to fall, it's speed increases, but at a rate less than g (it varies from g to 0 when (if) it reaches terminal velocity). With this lower value of acceleration, the time it takes to fall will be longer.