# Calculating vertical velocity component of a particle with mass, given the hit point of parabolic motion

Consider the following situation:

I have a particle with a given mass that at a given instant of time (let's say $t_{0}$) is placed at the system origin. The particle has a constant velocity component $V_{0x}$. I have to calculate the value of the vertical component ($V_{0y}$) to apply to that particle at time $t_{0}$, in order to provoke a parabolic motion that makes the particle hit the point $P$.

The data I know are the following:

• $P$ coordinates
• $dx$
• $V_{0x}$
• $t$ (that can be computed from $dx$ and $V_{0x}$)
• $m$ (the mass of the particle)

From the data above, is possible to calculate the value of $V_{0y}$? If yes then give some hints to solve it?

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This is easier than you think, because you're given the constant horizontal velocity $V_{0x}$, and as you say this allows you to calculate the time of flight $t$. So all you have to do is calculate the time for a particle with initial velocity $V_{0y}$ to either rise to a height $P_y$ in a time $t$, or more likely rise to a maximum then fall to a height $P_y$. Depending on the initial conditions both trajectories are possible.

The height of the ascending particle is simply given by one of the SUVAT equations:

$$s = ut + \frac{1}{2}at^2$$

where the initial velocity $u$ is $V_{0y}$, the acceleration $a$ is -9.81m/sec$^2$ (note the minus sign) and $s$ is $P_y$. I don't think I can say any more without spiling your fun :-).

Response to comment:

The site has very strict rules against doing people's homework problems for them but since you're 26 and an active Stack Overflow member I'm guessing this isn't homework and is related to some computing problem, so I'll go ahead. If the moderators object we may both end up in trouble :-)

We're given $P_x$ and $V_{0x}$ so the time of flight is just:

$$t = \frac{P_x}{V_{0x}}$$

Using the SUVAT equation I gave for $s$ above, and noting that $u$ is $V_{0y}$ and $s$ is $P_y$ we get, and substituting for $t$ we get:

$$P_y = V_{0y} \frac{P_x}{V_{0x}} - \frac{g}{2} \frac{P_x^2}{V_{0x}^2}$$

where I've used $g$ for the acceleration due to gravity (9.81m/sec$^2$). What we want is an equation for $V_{0y}$ and to get this we rearrange the formula to give:

$$V_{0y} = \frac{V_{0x} P_y}{P_x} + \frac{g}{2} \frac{P_x}{V_{0x}}$$

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well. first of all thank you for having put me in the right direction. I'm still trying to figure out how to complete your solution. As you can imagine, I don't know anything about physics. – Heisenbug Jan 8 '13 at 16:00
Really I can't get it. Could you kindly suggest me something more? – Heisenbug Jan 8 '13 at 16:39
thanks for the answer. As you guessed this is not homework (question has been retagged). I'm actually modifying some physic related C# code. Btw I already did those substitution you did above (thanks anyway for confirming that and for guided me in the right direction). I guess I have a bug somewhere else in the code, I'll try to find it since the calculus seem correct. – Heisenbug Jan 8 '13 at 20:13
As you said, depending from the initial conditions both ascending and descending trajectories are possible. Is there anyway I can get the right value of V0y only for the ascending one? – Heisenbug Jan 9 '13 at 16:59
Actually I got this wrong. Only one trajectory is possible, and whether it's ascending or descending depends on the initial conditions. In principle you can tell what type the trajectory is by calculating the $y$ velocity at the moment of impact: $V_y = V_{0y} - gt$. If $V_y$ is positive it's ascending and if $V_y$ is negative it's descending. – John Rennie Jan 9 '13 at 17:14

If this is motion under a gravitational field (i.e force is only along $-\hat{y}$, then you can derive the trajectory equation quite easily. It turns out to be:

$y=x\tan{\alpha}-\dfrac{gx^2}{2u^2}\cos^2{\alpha}$

To solve such a problem, you have to analyze motion along each axis separately and eliminate time from the equations of motions. For example, along the $Y$ axis we have:

$y=(v_0 \sin{\alpha})t-\dfrac{gt^2}{2}$

Along the X-axis we have:

$x=(v_0\cos{\alpha})t$ {No force along the X axis}

Now eliminate t from both equations.

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The answer is easily obtained by considering the free fall from a height h from the ground.The reversal of this motion corresponds to the motion of a projectile launched vertically upwards, that is, to the case V0x=0.

From the energy conservation equation: mgh = (mV^2)/2. So, V = sqrt(2gh). Here, V=V0y and h=Py. When V has a value that corresponds to the height Py,the projectile just reaches the height Py, reverses its motion and retraces its path. The positive and negative values of V correspond to the cases of opposite (upward and downward) directions of motion.

Non zero values of V0x make the path parabolic. The horizontal motion and vertical motion are independent and one does not effect the other. For values of V0y greater than that which corresponds to Py, the highest point of the path (apex of the parabola) is at a level higher than Py This case corresponds to the one shown in the daigram you gave), and for values of V0y lower than that which corresponds to Py, the projectile never reaches the height Py.

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