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This question already has an answer here:

My textbook says that a ball dropped vertically and a ball thrown sideways will not only both land simultaneously but their height will be corresponding for the entire fall, as shown in a diagram which has a ball falling vertically and a ball with an arch landing simultaneously.

This has really struck me as it feels intuitive that the ball dropped vertically would land faster, perhaps due to it traveling a shorter distance.

When I tried to think about it for myself, I came to the thought that perhaps it was due to gravity constantly pulling both of the balls, however would the sideways velocity not slow it downwards speed? For eg wouldn't a bullet take longer to land than a dropped bullet as it is traveling straight?

The image of the sideways ball

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marked as duplicate by sammy gerbil, Chris, Kyle Kanos, Jon Custer, Cosmas Zachos Feb 15 '18 at 16:23

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ as shown in a diagram can you show the diagram? $\endgroup$ – Wrichik Basu Feb 8 '18 at 6:55
  • $\begingroup$ oh yes, sorry, I have added it now. It is not the exact same one, but extremely similar and conveys the same ideas $\endgroup$ – John Hon Feb 8 '18 at 7:11
  • $\begingroup$ Then my answer would hold absolutely true. See if it helps you. $\endgroup$ – Wrichik Basu Feb 8 '18 at 7:12
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    $\begingroup$ Obligatory Mythbusters links. $\endgroup$ – Qmechanic Feb 8 '18 at 8:58
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    $\begingroup$ Possible duplicate of 2 balls falling hit the ground at the same time $\endgroup$ – sammy gerbil Feb 12 '18 at 0:30
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You've got it right when you say it's "due to gravity constantly pulling both of the balls" and NOTHING else pulling on the balls. Since gravity only acts in the vertical direction, what the balls are doing in the horizontal direction doesn't matter. Just remember that the thrown ball has to be thrown EXACTLY horizontally, and we are ignoring air resistance. Another answer to the title question could be "Why wouldn't they land at the same time?" You guessed that maybe the sideways velocity would slow the downward speed. Nope. That's the point of these types of physics problems. Gravity will affect a horizontally released object the Same way as one released when stationary. Same will happen with a bullet. Same for someone running off a cliff vs. walking off. You don't float in the air momentarily like the coyote and the roadrunner! Gravity starts acting immediately. So even a bullet is never "traveling straight"

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Imagine you and your friend are standing in front of each other. Your friend drops a ball. You look at your friend, and the ball takes a time $t$ to reach the ground.

Now imagine that instead of standing still, you are in a car travelling past your friend. The car is travelling at a velocity $v$. From your perspective, both your friend and the ball they're holding are travelling at velocity $v$. (From your perspective, you are travelling at $0$ velocity.)

Your friend then drops the ball. How long does it take the ball to reach the ground?

It seems preposterous that the ball would fall faster or slower just because you got in a car to look at it. The car ought to have no effect on the ball. Therefore, the ball still takes the same time $t$ to reach the ground.

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    $\begingroup$ In fact, imagine YOU are dropping an object from a moving vehicle. This might help intuitively. $\endgroup$ – Tom B. Feb 8 '18 at 7:50
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You say that the dropped ball should land sooner, as it travels a shorter distance. This reasoning would work if both balls were travelling at the same speed, which they are not.

The dropped ball has no extra force exerted on it by the dropper, and moves only due to the force of gravity, acting directly downwards along the y-axis.

The thrown ball, however, has had extra force exerted on it, in addition to that of gravity. In this idealized diagram, all the extra force is acting along the x-axis. Its y-axis velocity, the speed at which it moves downwards, is determined purely by gravity, just as in the case of the dropped ball, as none of the throwing force is directed downwards.

Both the balls have the same velocity along the y-axis and therefore should reach the ground at the same time.

(There is no reason that moving faster in one dimension should cause an object to move slower in another dimension, as you alluded to suspecting in your last paragraph. Using your example, a bullet shot downwards would reach the ground more quickly than if dropped, but this is because extra force has been exerted on it in the downwards direction by the gun. If the bullet was shot parallel to the ground, it would take as long to land as a bullet simply dropped (unless it was particularly aerodynamic and its movement generated lift, keeping it afloat longer).)

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Consider this situation where you have two planes travelling towards each other at some speed $v$, and as they pass each other both drop a ball: (the usual caveats apply: we're neglecting air resistance and any aerodynamic properties of the balls):

Dropped object

Hopefully it should be obvious that both balls will hit the ground at the same time because the situation is symmetric.

But now consider what the pilot of the plane on the left sees. As far as that pilot is concerned he drops his ball (the red ball) straight down so it falls in a straight line to the ground:

Dropped objects

But from his perspective the ball dropped by the other plane (the green ball) is travelling sideways at a speed $2v$ so it falls in a curve.

And the pilot of the plane on the right sees exactly the opposite. From his perspective he drops his (green) ball straight down while now it's the red ball that is moving sideways at a speed $2v$ and falls in a curve.

But we know that both balls touch the ground at the same time because the physicist watching from the ground sees both falls touch the ground at the same time. And that means the horizontal velocity cannot affect the vertical motion of the balls.

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  • $\begingroup$ This answer could be improved by noting whether you are considering aerodynamic effects on the ball or not. For why this matters, see related question physics.stackexchange.com/questions/491600/… . Of course if you are considering aerodynamic effects on ball then you need to state if there is any wind or not. $\endgroup$ – quiet flyer Jul 15 at 16:24
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It has to do with the displacement, not with distance.

A ball thrown sideways covers a greater distance, but the vertical displacement is the same as that of a ball thrown vertically.

The equation of motion is: $$h = \frac{1}{2} g t^2$$ where $h$ is the displacement and not the distance.

In both cases the value of displacement will be the same, and the initial velocity in the vertical direction will also be the same ($=0$)(take vertical component in case of a ball thrown sideways). That's why the time of fall would be the same.

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  • $\begingroup$ I feel like you meant to say "vertical displacement, not with total distance" in the first sentence of your answer, and "h is the vertical displacement" later, no? $\endgroup$ – Tom B. Feb 12 '18 at 18:58
  • $\begingroup$ Of course, everything is with displacement and not distance. $\endgroup$ – Wrichik Basu Feb 12 '18 at 19:32
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This is something that seemed mysterious to me for decades. It still does seem a little wondrous me if I am honest with myself but ultimately, it is simply an experimental fact that you can decompose a body's motion into separate, one-dimensional motions of its three, Cartesian position components and analyze them separately, applying Newton's second law using the component of the nett force on the body in each direction separately, and you'll get the right answer. The three separate motions really are independent of each other, as long as we decompose the force correctly into its Cartesian components at all times. So, in both cases, the acceleration of either body in the downwards direction is the same. the vertical position of both bodies is the same at the outset. Accordingly, the time to traverse the same vertical distance by both bodies is the same according to the above experimental fact.

Although this is a wholly experimental fact, we can refine its statement into more concise and mathematically powerful language by saying that force acts linearly, relative to the vector space of velocities, on a particle's motion state and that velocity is a vector, in the mathematician's sense. Mathematically, a vector is something which belongs to a vector aka linear space; that is, a set of objects which can all be represented as a linear superposition of basis objects. Position in classical mechanics on a flat space is such a vector: something decomposable into a superposition of the basis displacements from a reference point. Simple theorems tell us that we can do the decomposition into any basis and the results will always be consistent and that moreover all basisses have the same number (the dimension) of basis vectors (the wonted plural is bases, but looks a bit confusing since it looks the same as baseball and military bases but has a different, long-e and stressed sounding second syllable). The superposition weights are our position co-ordinates. Likewise for velocity and acceleration. When we say that an operator acts linearly, we mean that its action on a superposition of vectors is the same superposition of the its action on the individual vectors.

Which, in this case, is simply a fancy way of saying that we can decompose the analysis of the operator's action as we do for the force acting on a point body.

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the horizontal component of velocity of the projected object is zero in both the cases. Thus time of flight is same in both the cases because every component of motion is independent of the component perpendicular to it.

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    $\begingroup$ If horizontal component is 0, the green ball would have landed just where it was thrown. There would have been no horizontal range, like the blue one. $\endgroup$ – Wrichik Basu Feb 8 '18 at 7:17

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