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.
[Physics] Does gravity slow down a horizontally thrown baseball
dragnewtonian-gravitynewtonian-mechanicsprojectile
Related Solutions
Hint: I think the issue here is that the motion here is not symmetric. If it starts out with a speed $u$, it is not necessary that it will have the same speed when it reaches the bottom - because the acceleration is not the same in both the cases.
In the case without air resistance, it is valid to write $t=\frac{u}{g}+\frac{u}{g}$, because the particle goes from $u$ to $0$ and then $0$ to $u$, which doesn't happen in this case.
Try studying the problem by taking into account the distance the projectile travels - because the distance it travels up is always the distance it'll travel down.
The key here is the fact that the ball going down is 'fired straight down'. In other words it isn't just dropping with the force of gravity, but with an initial velocity already given to it. Due to the SUVAT equation $v=u+at$ we can see that this initial velocity is added to what it would've been if it was just left to fall. In the answer you have given it says "They each start with the same kinetic energy". This means that each ball is fired with the same initial velocity. Let's work through this mathematically. The height that the ball fired up reaches is $$s=\frac{u^2}{2g}$$ Taking $g=10$ for simplicity, we can say that the ball reaches a height of $0.05u^2$. Using the equation $$v^2=u^2+2as$$ we can calculate it's final velocity, which comes out to be $$v^2=0+2*10*(0.05u^2+h)$$ or $$v=\sqrt{u^2+20h}$$
Let's now think about the ball fired downwards. This is going to have a final speed of $$v^2=u^2+20h$$ or $$v=\sqrt{u^2+20h}$$
Now let's do the velocity for the final ball, projected horizontally. We can calculate it's vertical velocity by using $$v^2=u^2+2ah$$ and from this we get $$v_{vert}=\sqrt{20h}$$ As there is nothing happening to the horizontal velocity it is still just $u$ at the end, so for the ball fired horizontally we have $$v_{vert}=\sqrt{20h}$$ $$v_{hor}=u$$
Here we need to calculate the magnitude of the velocity, $\|v\|$, by taking $$v=\sqrt{v_{vert}^2+v_{hor}^2}$$ We can see here that if we do this we end up with $$v=\sqrt{u^2+20h}$$
So all balls end up with the same speed at the end. It is much easier to do this sort of thing as a change in energy, however this way is also an effective method of proving it.
Hope this helps :)
Best Answer
No, gravitational acceleration is vertically downwards. It does not affect the horizontal velocity of any projectile.