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

Question

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?

Answer 1

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:

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

Answer 2

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.