# Is it possible to throw an object faster but over the same distance?

Is it possible to throw say a tennis ball at $1ms^{-1}$ for 20m and then be able to throw the same ball at $2ms^{-1}$ for 20m at the same angle? That is throwing the ball in a traditional curve withe surface between the landing point and the take off point being flat.

Logic screams that this is not possible if so what is the relation between the speed and the distance (I mean what says how big the distance will be depending on how fast the ball is thrown).

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Yes it is. Try at angle $\pi / 2$ wrt ground! Then the distance between take off and landing point for both the cases will be same. :) –  Pratik Deoghare Nov 20 '10 at 8:06

The relationship is between speed, distance, and the angle thrown. The distance the ball travels before coming back to the same height is further if you throw it faster, but less if you throw it at a lower angle (up to about 45 degrees). By adjusting both the speed and angle of your throw, you can have two throws that go the same distance at different speeds.

However, neither 2m/s nor 1m/s is fast enough to throw something 20m. There is a maximum distance you can throw for a given velocity. This distance depends on the square of the velocity, because if you throw something faster at a given angle it will be up in the air for more time and travel faster during that time. It also depends inversely on gravity's strength, because if gravity becomes stronger the time in the air is less. $g$ is about 10m/s^2, so for a throw of 2m/s $v^2/g = .4m$ is nowhere close to the desired 20m throw.

The full expression for the distance traveled, neglecting air resistance, is

$\sin(2\theta)\frac{v^2}{g}$

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Like you said by adjusting speed and angle you can have different throws with the same distance, however the OP specified with the same angle, in that case then you can't, right? –  teto Dec 20 '10 at 0:19
Correct. If you chck the revision history of the question, though, you'll see the OP added in the part about same angle after I answered. –  Mark Eichenlaub Dec 20 '10 at 1:26

The total distance traveled by a ball launched with velocity $v$ at an angle $\theta$ from the horizontal to reach the save vertical level is

$$\Delta x=v^2\sin(2\theta)/g$$

That comes from

\begin{align*} \Delta x & =(v\,\cos\theta)\, t\\ \Delta y & =(v\,\sin\theta)\, t-\frac{1}{2}\, g\, t^{2}\end{align*}

solved for $t$ when $\Delta y = 0$ as $t=2\, (v\,\sin\theta)/g$ and ignoring the trivial solution of $t=0$.

So the answer is no because for the same $\Delta x$ when $v$ changes so does $\theta$ has to change.

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In reality you could do that by throwing curved balls. You curve the slower ball in one direction so it travels further and you curve the faster one in the opposite direction. You can see that in table tennis.

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Or in baseball. Even in soccer. –  Raskolnikov Dec 3 '10 at 13:30
+1: this is the only answer that gets near correctness without saying it's impossible. Since the problem without rotation has only one unique solution, a new variable needs to be added, angular momentum, which could allow different distances at the same angle and velocity. If the path of the ball is the same, I don't know and that would take some vector math... –  rubenvb Dec 19 '10 at 11:56

It's possible to have two angles with the same initial velocity that carry the ball the same distance.

You can play here.

I didn't run through the calculations for your particular example, but you can do that easily enough yourself if you want.

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If you throw a ball with an angle $\theta$ and initial velocity $v_1$, the ball will hit the ground at a distance given by

$d_1 = \frac{v_1^2}{g}sin(2\theta)$

Similarly, if you throw it again with a different velocity $v_2$, the ball will hit the ground at

$d_2 = \frac{v_2^2}{g}sin(2\theta)$

What you are asking is if there is such a $\theta$ for which $d_1 = d_2$ while $v_1 \neq v_2$.

Let's see:

$d_1 = d_2 \implies \frac{v_1^2}{g}sin(2\theta) = \frac{v_2^2}{g}sin(2\theta)$

or

$v_1^2 sin(2\theta) = v_2^2 sin(2\theta)$

Now, if $sin(2\theta) \neq 0$ this leads to $v_1^2 = v_2^2$ which is never true for positive velocities.

Therefore,

$sin(2\theta) = 0$

and this gives the following solutions: $0, \frac{\pi}{2}, \pi...$.

In other words, if the the angle is 0 or 180 degrees, then the ball will hit the ground instantaneously in both cases. If the angle is 90 degrees, the ball will simply go up and fall back on the same spot in both cases.

Note: this does not take into account air resistance, variations in the acceleration of gravity due to distance and the rotation of the Earth.

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I suppose you could reproduce the exact trajectory (i.e. same angle and distance) by changing gravity, or more physically realistic, using an electric charge on the ball and uniform background E field.

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