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For a projectile motion in a $2D$ plane, if the path of trajectory is a parabola with initial angle of projection to be $\theta$, explain me how to derive the time of flight, horizontal distance traveled by the object after a particular time (displacement of the object), and the range (total horizontal distance traveled by the object).

I do understand that the $x$ component of initial velocity is $u \cos \theta$ and $y$ component is $u \sin \theta$. But this $\theta$ and $u$ is continuously changing, isn't it? So, if we consider the equation $S=u_xt+\frac{1}{2}gt^2$, then it gives the displacement or the total path of the trajectory? Do explain the reason.

Please explain everything using calculus (integration) as much as possible and usage of equation of parabola would also help.

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  • $\begingroup$ If acceleration is 0 then the projectile will follow a straight path. It will be the same thing as throwing a projectile in deep space. It will follow a straight path throughout its motion. $\endgroup$ – Mitchell May 18 '17 at 5:11
  • $\begingroup$ @BhavyaSharma edited. $\endgroup$ – Mathejunior May 18 '17 at 5:17
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    $\begingroup$ You should consider adding in your efforts towards the derivation. Otherwise your post will be closed. Show your attempt and failures to derive the equations. $\endgroup$ – Mitchell May 18 '17 at 5:19
  • $\begingroup$ @BhavyaSharma The proof is given in the book which I'm studying. But I cannot understand it well. Hence I posted it here. $\endgroup$ – Mathejunior May 18 '17 at 5:24
  • $\begingroup$ What does "total range" mean? $\endgroup$ – Steeven May 18 '17 at 5:44
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We always start from the four kinematic motion equations:

$$s=s_0+v_0t+\frac 12 at^2\\ s=s_0+\frac 12(v_0+v) t\\ v=v_0+at\\ v^2=v_0^2+ 2a(s-s_0)$$

These equations count along any path. The $s$ can be replaced with a $x$ when looking at a horizontal path (the speeds and acceleration shall then also be for this horizontal direction), and with a $y$ for a vertical path, which in total gives you up to 8 equations, if so many should be needed. They are derived solely from the basic definitions $v=dx/dt$ and $a=dv/dt$ in the case of constant acceleration $a$, which is the case in freefall here on Earth.

You have identified the start speeds horizontally and vertically: $v_0|_x=v_0\cos \theta$ and $v_0|_y=v_0\sin \theta$, which you can plug into the equations when necessary.

But this θ and u is continuously changing, isn't it?

No, the throw angle $\theta$ and throw speed $v_0$ will not change during the flight. This is just how the throw was done, and how it flies doesn't alter how it started - these parameters will be different for other throws but do not change continuously during the flight.

  • If you need the time $t$, you just figure out which other parameters you know in your specific case and try to rearrange and isolate $t$ in any of the above equations. You might need to insert one equation into another and a lot of algebra can be needed in some cases. But you will almost always have enough equations to take from (you have 8 to choose from).

  • If you need the horizontal distance travelled after some time, then you are looking for $x$. Just remember to plug in the correct time $t$ and then the $x$ you get will fit to this exact time. Same procedure: Consider a horizontal path and input the horizontal start speed $v_0|_x$, horizontal speed $v_x$, horizontal acceleration $a_x$ and call the distance $s$ for $x$ instead to make it easier to overview.

    • In the horizontal case, rememer that there is no horizontal acceleration in a free-fall, so $a_x=0$.
  • If you need the total horizontal distance travelled, then do the same, but plug in the total time as $t$ instead (the time when the object lands again). You might have to calculate this time first.

if we consider the equation $S=u_xt+\frac 12gt^2$, then it gives the displacement or the total path of the trajectory?

I am not sure this is a correct formula for the trajectory length (the length of the flight path).

Firstly, remember that here on Earth, while there is no horizontal acceleration $a_x=0$, there is a vertical acceleration, which has a known value: $a_y=g$ (pointing downwards, so remember a proper sign). Therefore a $g$ can appear.

But I would try to find the actual trajectory length by first making a trajectory equation (see here for example; go down to the "Parabolic Trajectory" headline) and then to use the length formula on it (see here). Those expressions are:

$$\begin{align} \text{Trajectory equation: }\quad &y=\tan \theta x-\frac g{2 v_0^2 \cos^2 \theta}x^2\\ \text{Path length formula: }\quad &Length=\int_a^b\sqrt{1+(y')^2} \;dx \end{align}$$

What you already have here does not look right to me, but I can't be sure (since I don't remember this expression by heart) without going through these steps to obtain it.

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$\vec{s}=\vec{u}t+\frac{1}{2}\vec{a}t^2$

The above relation can be used for $x$-axis or y-axis separately, you just have to keep the vectors in your mind. This relation can be broken down into two components :

$s_x \hat{i} + s_y \hat{j}=(u\cos{\theta} \; \hat{i} + u\sin{\theta}\; \hat{j}) t + \frac{1}{2}(0\;\hat{i}+g\;(-\hat{j}))t^2$.

Things will be more clear if we separately compare the displacements along the two axes.

For $x$-axis :

$s_x =u\cos{\theta}\;t$ $\tag1$

The direction of all the vectors is the same and the projectile will only travel in a particular direction $(+\hat{i})$, so we can neglect the unit vectors and focus on the magnitude of the vectors. This is only for $x$-axis.

There will be no change in the horizontal component of the velocity, since there is no acceleration along $x$-axis to bring about that change. You increase the time, the displacement along $x$-axis will keep on increasing.

For $y$-axis :

$s_y.\hat{j}=u\sin{\theta}\;(\hat{j}) t-\frac{1}{2}g\;(\hat{j})t^2$. $\tag2$

In case of $y$-axis we have to consider the direction as well as the magnitude of the vectors, A projectile passes through the same height twice (except maximum height).

Range :

Range for a projectile is defined as the horizontal distance covered by the projectile in that time period (i.e., time of flight) in which the projectile remains in air (roughly speaking).

So, if you throw a projectile from the ground, the time of flight will be equal to the time that it takes for the projectile to attain the maximum height and again come back to the ground $(s_y=0)$.

And in this time, the horizontal distance covered by the projectile will be its range.

This time can only be found out by equation $(2)$, since we are considering displacement along $y$-axis to define the time of flight.

$s_y$ is $0$ at two instants, the initial instant when the projectile was thrown and the final instant when the projectile reaches the ground again.

We are interested only in the second case.

So, $T=\frac{2u\sin{\theta}}{g}$.

To find the range, you only have to put this value of time in equation $(1)$.

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