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I have a simple question: Does gravity slow down a horizontally thrown baseball?

Assuming when a baseball is thrown it has a vertical velocity as well, does it slow does the ball?

Any help is much appreciated.

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The second part of your question asks about the case when the ball is thrown with a non-zero vertical velocity. Let us consider the case where the initial vertical component of the velocity is in the upward direction. In that case, gravity still acts on the vertical velocity, and not on the horizontal velocity, but the magnitude of the velocity vector (speed) will decrease until the ball reaches the peak of its flight. Then, as the ball's velocity grows increasingly in the downward direction, the speed increases. – kleingordon Jul 30 '13 at 5:47
up vote 6 down vote accepted

(I can't quite comment on the previous post, so I'll have to write a new answer).

If we set the curvature of the Earth to be non-negligible in our problem, yes, gravity would slow the baseball down by an extremely tiny amount, but, if we exclude this case (which, again, I stress to be many orders of magnitude below anything considerable), then no, gravity itself does not slow the ball down since the force of attraction (the direction of the vector of acceleration) points exactly downwards and contributes nothing to the horizontal component.

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So the ball purely slows down due air friction? – user793468 Jul 29 '13 at 18:03
Yes, essentially. Since the curvature at this point is extremely tiny, pretty much 99.99999999% (and a few more, but this is just an order-of-magnitude estimate) of the contribution to the ball's horizontal slow-down is due to the air resistance. – Guillermo Angeris Jul 29 '13 at 18:09

No, gravitational acceleration is vertically downwards. It does not affect the horizontal velocity of any projectile.

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So the ball purely slows down due air friction? – user793468 Jul 29 '13 at 18:02
Yes, as long as the ball is considerably smaller than the earth. – user80551 Jul 29 '13 at 18:03

I'll say no the velocity is constant.

At $t=0$ when you throw the ball, you give it a horizontal velocity $\vec{v_0}$.

At a certaint time $t$, we have according to the second Newton's law, since the ball is affected only by its gravity as you said (so we neglect any friction):$$\vec{P}=m\vec{a}$$ where $\vec{P}$ is the gravity attraction on the ball and $\vec{a}$ is the ball's acceleration and $m$ the mass of the ball.

In the cartesian coordinate system, where $\vec{i}=\dfrac{1}{v_0}\vec{v_0}$ (which means that $\vec{i}$ has the same direction and orientation and $||\vec{i}||=1$).

We have here $P_x=ma_x=m\dfrac{dv_x}{dt}$

While $\vec{P}$ is perpendicular to $\vec{i}$ because $\vec{i}$ is horizontal and $\vec{P}$ is vertical, then $P_x=0$ which means that $\dfrac{dv_x}{dt}=0$ and so $v_x=cte=v_0$

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Gravity only acts on the vertical component, not the horizontal component. The force of gravity speeds an object down on decent as gravity makes the ball fall to earth.

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protected by Qmechanic May 22 at 21:02

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