An object thrown vertically upwards into the air takes shorter time to reach its peak height than it does to?

Question

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,

Answer 1

I am not entirely sure exactly what your asking...

If you are referring to total time, then yes it will take less time for the object to reach its peak before returning to the level of the thrower.  Say the object is thrown up at time t=0.  If it reaches its peak at t=5 seconds, then it will reach the thrower at t=10s.  The first 5 seconds the object is decelerating upwards, and the last 5 seconds it is accelerating downwards.

If you are asking whether or not the time to reach the peak takes less time than the time to reach the thrower from the peak, the answer is that they will take the same amount of time.

This can be shown with the equation V=Vi + at. Where:

Vi=initial velocity, a=acceleartion (negative if decelerating), and t=time.

You know that at the peak of the arc, the velocity of the object will be 0.  Assume that you threw the ball upwards with an initial velocity of 25 m/s, we will get:

0=25+(-9.81m/s^2)t or t=2.5484 seconds

Subbing this into the distance equation: d=Vit+0.5at^2

d=25(2.5484)+0.5(-9.81)(2.5484)^2 or d = 31.8553 m to reach the peak (neglecting air resistance).

Using this to determine the time to fall from the peak:

31.8553=0(t)+0.5(9.81)t^2 or t=2.5484 seconds.

This is exactly the same as the time it took to reach the peak.

Basically, the same acceleration or deceleration is applied to the object, making the travel up the mirrior image of the travel downwards.  Air resistance, although neglected here, woudl not change the outcome as there will be equal air resistance in both cases (assuming that the balls total height is small and aire pressure remains constant throughout its trip)

EDIT:

Ok after re-reading the question, you are right there would be a slight difference in the times if air resistance was included.  Lets say that the air resistance (as you mentioned is a force) causes a deceleration of 1 m/s^2.  When the ball is thrown upwards it will decelerate at 9.81 m/s^2 + 1 so a total of 10.81 m/s^2.  On the way down, it will be accelerating at 9.81 m/s^2, but still decelerating at 1 m/s^2 for a total acceleration of 8.81 m/s^2 so it will take slightly longer to hit the ground.  This can be proved with a balloon which greatly eggagerates the effects of air resistance.  You can throw a balloon up fairly quickly, but it will always take a long time to fall to the ground.  Thus, as you mentioned, if you include air resistance the time to get to the peak will be shorter than the time to decend back down to the thrower.

Good question!

Answer 2

Theoretically, it should take the same time to reach the peak height and to fall back down. In actuality, this is false. You are correct in saying that it has gravity and air resistance slowing it down on the way up and only air resistance working against it on the way down.

Answer 3

EDIT

Thanks for sorting that out JerkFace. The balloon was an great example.

Answer 4

umm, 42?