# When we r considering a ball that is thrown at 45 angle, with the air

when we r considering a ball that is thrown at 45 angle, with the air drag present, why does it take longer to fall back after reaching its highest point???

The air exerts a sort of friction force on the ball, taking away some of its mechanical energy and converting it into thermal energy.

So as the ball travels along its trajectory, some of its kinetic energy is being converted into thermal energy. Hence the ball's velocity is decreasing, which means it takes longer to reach the ground than it did to reach its highest point.

Drakkith
Staff Emeritus

Consider a more extreme example. I throw a feather straight up. It takes about 1 second for it to reach maximum height, but several seconds to fall back down to the height I released it at.

an explanation stated that when its going up, the air resistance is in the same direction as the weight(mg) acting, and when its going down, the air resistance is in the opposite direction as that of the of the weight(mg)..thus in the second half of the motion (that is when its coming down) the speed is lesser..
what i dont get is that when its going up shouldnt the speed be lesser since the resistance and weight BOTH r acting to oppose the motion.........?

Danger
Gold Member

I think that the operative factor here is that upon descent the only forces acting upon the ball or bullet or whatever are gravity and air resistance. That same projectile typically undergoes several g's upon launch.

phinds
Gold Member

I think that the operative factor here is that upon descent the only forces acting upon the ball or bullet or whatever are gravity and air resistance. That same projectile typically undergoes several g's upon launch.

So you reckon that if it was happening in a vacuum it would take longer to come down than to go up? And that it would git the earth will less G force than that with which it was launched? It certainly SOUNDS like you believe that.

Danger
Gold Member

So you reckon that if it was happening in a vacuum it would take longer to come down than to go up? And that it would git the earth will less G force than that with which it was launched?

'No' and 'yes', in that order. (I assume that the last sentence is supposed to have the words 'hit' and 'with' in it.)
As I pointed out, one of the factors is air resistance. The vacuum state would eliminate that. In normality, though, the object can be accelerated at any rate that doesn't achieve escape speed, but will always cease accelerating upon return when it hits terminal speed. Of course, the initial acceleration ends upon release from the launcher whereas gravity is constant, but the launcher power is ungoverned within engineering limits.

phinds
Gold Member

'No' and 'yes', in that order. (I assume that the last sentence is supposed to have the words 'hit' and 'with' in it.)
As I pointed out, one of the factors is air resistance. The vacuum state would eliminate that. In normality, though, the object can be accelerated at any rate that doesn't achieve escape speed, but will always cease accelerating upon return when it hits terminal speed. Of course, the initial acceleration ends upon release from the launcher whereas gravity is constant, but the launcher power is ungoverned within engineering limits.

I'm clearly missing something since it appears that you are contradicting yourself. You say that air resistance / terminal velocity is why it would not hit the ground (in an atmosphere) with the same G force with which it was ejected, and then you also say (rightly) that the vacuum state would eliminate that. Them implication to me of all that is that, clearly, it would hit the ground with exactly the same G force as that with which it was ejected, but you say that's not true. I don't follow your logic

Danger
Gold Member

Okay... I can see where there might be some misunderstanding due to my choice of words.
Try this:
If you fire a rifle straight up, the bullet leaves the muzzle at several thousand g's. When it reaches the apex of it's trajectory, it essentially stops and begins to fall (not exactly, of course, but close enough). On the way back down, it accelerates at 1 g. The terminal speed that it reaches will be nowhere near the muzzle velocity of the firearm.

phinds
Gold Member

Okay... I can see where there might be some misunderstanding due to my choice of words.
Try this:
If you fire a rifle straight up, the bullet leaves the muzzle at several thousand g's. When it reaches the apex of it's trajectory, it essentially stops and begins to fall (not exactly, of course, but close enough). On the way back down, it accelerates at 1 g. The terminal speed that it reaches will be nowhere near the muzzle velocity of the firearm.

NO, by your definition, we are talking here about a VACUUM. There is no terminal velocity Why would the bullet not have exactly the same velocity when it comes back as when it left?

Danger
Gold Member

I very specifically pointed out that I am not referring to a vacuum. You, in fact, were the one who introduced that concept. I'm dealing with real-life situations, in which a vacuum plays no part. I have no idea of how you got so turned around on this.

phinds
Gold Member

I very specifically pointed out that I am not referring to a vacuum. You, in fact, were the one who introduced that concept. I'm dealing with real-life situations, in which a vacuum plays no part. I have no idea of how you got so turned around on this.

OOPS. You're right. Don't know how I got it in my mind that YOU brought up vacuum when clearly I did that.

Danger
Gold Member

No sweat, pal.

why several 1000 g's in the beginning.....................??

Drakkith
Staff Emeritus

why several 1000 g's in the beginning.....................??

Because it accelerates in the barrel from stationary to around 2000 ft per second or so in milliseconds.

Don't forget that air resistance varies with velocity so it is much greater when the object is traveling very fast so it will slow the object very quickly at the beginning.

phinds
Gold Member

Don't forget that air resistance varies with velocity so it is much greater when the object is traveling very fast so it will slow the object very quickly at the beginning.

Yes, you are correct, but it has no effect on the vertical component of velocity so really doesn't matter to this discussion.

when we r considering a ball that is thrown at 45 angle, with the air drag present, why does it take longer to fall back after reaching its highest point???

and a second quote Yes, you are correct, but it has no effect on the vertical component of velocity so really doesn't matter to this discussion.

I'm posting information regarding the first question of the thread which concerned an object being thrown through the atmosphere rather than through a vacuum. I can only assume the second quote refers to a vacuum because most certainly air friction slows both components of the velocity as the ball moves through air as it slows the actual velocity. So my response is definitely relevant.

the faster the object moves through the air, the more molecules it will hit and the more it will be slowed, lose energy to friction and so on. It will reach it's lowest upward velocity at the maximum height and from there, even though gravity is acting on it in the downward direction, air friction is acting on it in both the horizontal and vertical directions slowing it more and more in the horizontal direction and reducing the acceleration in the vertical direction. You can take video of the ball in the air and frame by frame it to see exactly what happens and to determine the velocity for any interval so you can see this is true. The horizontal velocity does not limit the time in the air though so we have to consider exactly what happens to the vertical velocity and assume the object falls the same distance vertically as it rose. ON the up trip both friction and gravity are acting downward and friction is the largest when the object is first thrown and you have friction in the horizontal direction but again this doesn't affect the time in the air, only how far horizontally the ball will travel. So greatest deceleration is at the beginning of the up trip but also greatest velocity. On the down trip gravity speeds the ball up and friction slows it. So to decide if the ball really does take longer to come down you have to decide if the velocity is totally symmetric in the vertical direction. The path is certainly not symmetric but what about the y velocity?

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phinds
Gold Member

netgypsy, you are obviously correct. I got sidetracked and was thinking that the entire conversation had been about horizontally fired bullets ... my bad.

This question had not yet been satisfactorily been discussed.

Consider two possible situations - one in which the launch velocity exceeds terminal velocity and one in which it does not.

Drakkith
Staff Emeritus

This question had not yet been satisfactorily been discussed.

Consider two possible situations - one in which the launch velocity exceeds terminal velocity and one in which it does not.

I don't see any difference in the two.

One looks for symmetry in this case because if the up trip and the down trip are identical (forget the horizontal as it has no affect on the time in the air) if all variables in the up and down just reverse, we should not have a time difference for the up and down trips.

Suppose you shoot a nerf ball directly upward with a launcher and it leaves the launcher at a speed significantly higher than terminal velocity. The ball will slow down very quickly until it reaches terminal velocity and from there of course, slow to a stop and accelerate toward the ground.

It now falls straight down. BUT on the down trip it is not possible for the ball to reach the launch speed because it has no internal motor or rocket engines so it will fall more and more rapidly until it reaches terminal velocity where the force due to air friction and the force due to gravity are equal, at which time it will fall at constant speed until it reaches the same elevation at which it was launched.

So one can see that the up and down trips are not symmetric in this case. The up trip has a faster velocity at the beginning of the free flight than it does at the same position on the down trip. So it does take longer for it to fall on the down trip just by intuition.

But what if you launch the ball vertically AT terminal velocity? How does this change the problem?

Drakkith
Staff Emeritus

It doesn't. The faster you fire the projectile upwards, the further it goes before reaching its peak. Terminal velocity has absolutely nothing to do with the projectile on its upward trajectory.

If it's fired faster than terminal velocity the downward velocities will not match the upward velocities so there is no symmetry. There is no way the object can go downward with a velocity greater than terminal velocity but it can go upward with a velocity greater than terminal velocity so the up trip and down trip don't take the same length of time even though they go the same vertical distance. You are correct that if you fire the object at a speed greater than terminal velocity it will go higher than it would if you didn't, just as you said. But that wasn't the question. The question was why does it take longer to fall than rise.

If you fire the object AT terminal velocity, what happens?

Drakkith
Staff Emeritus

I assume that the time is the same.