Translational Motion - probably a simple Q

[LetsGo352]

I have a simple Q. When you toss a ball up with an initial velocity, why does the ball the take the same amount of time to go back as it does back down? I understand all the calculations that are associated with this, but I just can't grasp the concept for some reason. I mean in one direction it's decelerating and in one direction it's accelerating. I just don't understand the concept - can someone clear this up for me? Thanks

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[matth87]

Think about it like this

Say you throw the ball up with a velocity of 10.

It decelerates from 10 to 0.

Then

It accelerates from 0 to 10.

Its just the reverse, so it takes the same time.

 

[G1SG2]

I have a simple Q. When you toss a ball up with an initial velocity, why does the ball the take the same amount of time to go back as it does back down? I understand all the calculations that are associated with this, but I just can't grasp the concept for some reason. I mean in one direction it's decelerating and in one direction it's accelerating. I just don't understand the concept - can someone clear this up for me? Thanks

In actuality, if one takes air resistance into account, the time up is a bit less than the time coming down since on the way up, air resistance acts in the same direction as gravity. On the way down, it oppose gravity. Thus, time up is less than time down. But if we ignore air resistance, all you have acting on the ball is gravity. Gravity slows the ball down as it goes up by 10 m/s^2, and also accelerates it as it falls by 10 m/s^2. The only force acting on the ball would be gravity, one the way up and on the way down. Think about it this way-if the ball, in either direction, were to have more force acting on it, it would accelerate/decelerate more, right? As from F=ma, we can see that a=F/m. A greater force would give a greater acceleration, resulting in a shorter time. But on the way up AND on the way down, the only force acting on the ball is gravity. It's the same amount of force. Thus, time has to be the same.

 

[LetsGo352]

Ok these posts cleared it up a lot. Thanks a lot.