# How do they derive these formulas? [closed]

I've been doing some practice problems and they use formulas such as

$$x=\frac{v_0^2\sin(2\theta)}{g}$$

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

Can somebody please inform me on the ways in which they derive these formulas?

• You should try to format your equations properly (so that we see mathematicak symbols rather than their names) and explain what are some of the symbols that you use. Based on my guess what is meant here: did you try to use conservation of energy and/or momentum? Mar 23, 2021 at 23:10
• @DanDan0101 Thanks. A ball kicked from ground level at an initial velocity of 60 m/s and an angle θ with ground reaches a horizontal distance of 200 meters. a) What is the size of angle θ? I tried solving this problem without the formula and I got it wrong so I would like to know in which scenerio do we use the formula. Take a look at the question yourself by clicking the following link problemsphysics.com/mechanics/projectile/… it is the fifth problem in the site. Mar 23, 2021 at 23:52
• I upvoted to cancel out the downvote, I think it's reasonable to want to understand how such equations are derived and appropriate for this site. @Ran Sh2, are you familiar with the standard kinematic equations? Mar 23, 2021 at 23:53
• @electronpusher The question as it stands is not suitable for this site as it just lists a couple of formulae and asks as to how one derives them. The OP should clarify what they understand about the background subject matter and at what point they are getting stuck in deriving these formulae, etc.
– user87745
Mar 24, 2021 at 1:03

Thank you to everybody who commented and tried helping, I found out the answer. For the $$\Delta x$$ formula you use $$v_{fy} = v_{0y} + ay \cdot t$$ and then you isolate $$t$$ and plug it into $$s_x = v_x \cdot t$$ then use pythagorean identities and you have the first formula!

For the next one use $$v_f^2 = v_0^2 + 2ad$$ then isolate $$y$$ granted that $$v_{fy} = 0$$. Then you have your second formula!

The first equation is for the projectile range, $$R$$, $$R=\frac{v_0^2\sin(2\theta)}{g}$$ This assumes that the projectile starts and ends at the same height.

We know that $$\Delta y = -\dfrac12 gt^2+v_0 \sin(\theta) t$$ where $$v_{0,\; y}=v_0\sin\theta$$. Since the projectile lands at the same height from which it was launched, we have $$\Delta y = 0$$, which means: $$0=-\dfrac12 gt^2 + v_0\sin(\theta) t\quad \implies 0=t\left(-\dfrac12gt+v_0\sin\theta\right).$$

The two times $$t$$ for which $$\Delta y=0$$ are $$t_1=0$$ and $$t_2=\dfrac{2v_0 \sin\theta}g$$. The trivial solution is $$t=0$$, we only care about the landing time, $$t_2=\dfrac{2v_0 \sin\theta}g$$.

Recall that the x-displacement is $$\Delta x = v_0\cos\theta \cdot t$$. We're after the range, so we get $$R=v_0\cos\theta \cdot t_2\implies R=v_0 \cos\theta \dfrac{2v_0 \sin\theta}g.$$

Recall the trig identity $$\sin2\theta \equiv 2\sin\theta\cos\theta$$. Applying it, we get,

$$R=\dfrac{v_0^2 \sin2\theta}{g}.$$

The second equation is for the max displacement of your projectile, $$\Delta y_\mathrm{max}=\frac{v^2\sin^2(\theta)}{2g}.$$

Easiest way to get it is to use the $$v_f^2 = v_0^2 +2a\Delta d$$ kinematic equation. Note that we are working in the y-direction, so this equation becomes $$v_{f, \ y}^2 = v_{0, \ y}^2 -2g\Delta y.$$

Note that the y-velocity at the max height is zero; that's what makes the height a max. We then have, $$2g\Delta y = v_{0, \ y}^2.$$

We know that $$v_{0, \ y} = v_0 \sin\theta$$, so we have

$$2g\Delta y = \left(v_0 \sin\theta \right)^2$$

which gives the max displacement and simplifies to $$\Delta y_\mathrm{max}=\frac{v^2\sin^2(\theta)}{2g}.$$

Nothing wrong with those two formulas.

The distance traveled by any particle along a direction can be written using the following two formulas. This you can find in any basic physics book.

\begin{align} S_{x} &=u_x*T ~~~~\text{when moving at constant velocity} \label{eq:01} \\ S_y &=u_y*t+\frac{1}{2}f*T^2 ~~~~\text{when moving with acceleration} \label{eq:02} \\ \end{align} In your problem, for the horizontal direction moion, \begin{align} S_{x} &=u_x*t \label{eq:03} \\ S_{x} &=v_0*cos(\theta)*T \\ T &= \frac{S_{x}}{v_0cos(\theta)} \\ \end{align} For the verticle motion, $$S_y$$ is zero as the particle starts from the ground then goes up certain distance and then comes back to ground, net verticle distance is zero. \begin{align} S_y &=u_y*T+\frac{1}{2}f*T^2 \label{eq:04} \\ 0 &= v_0sin(\theta)*T + \frac{1}{2}*(-g)*T^2 \\ ~~~~ & \text{Note: g is acting in -ve y direction, that's why -ve sign}\\ v_0sin(\theta)*T &= \frac{1}{2}*g*T^2 \\ T &= \frac{2v_0sin(\theta)}{g} \\ \end{align} Equate those two T's you will get your first equation. \begin{align} \frac{S_{x}}{v_0cos(\theta)} &= \frac{2v_0sin(\theta)}{g} \\ S_x &= \frac{v_0^2sin(2\theta)}{g}\\ \end{align}

$$\begin{equation} \boxed{x = \frac{v_0^2sin(2\theta)}{g}} \end{equation}$$

$$\Delta{y}$$ is the maximum height. At max verticle height $$v_y=0$$. Use this condition to get the time it takes to reach maximum height.

\begin{align} v_y &= v_0sin(\theta) -g*T' ~~~\text{basic formula: v=u+ft}\\ 0 &= v_0sin(\theta) -g*T' \\ T' &= \frac{v_0sin(\theta)}{g} \\ \end{align} Then maximum distance traveled in the verticle direction becomes, \begin{align} \Delta{y} &= v_0sin(\theta)*T' -\frac{1}{2}g*{T'}^2 \\ &= v_0sin(\theta)*\left[\frac{v_0sin(\theta)}{g} \right] -\frac{1}{2}g*{\left[ \frac{v_0sin(\theta)}{g} \right] }^2 \\ &= \frac{v_0^2sin^2(\theta)}{2g} \\ \end{align} $$\begin{equation} \boxed{\Delta{y} = \frac{v_0^2sin^2(\theta)}{2g}} \end{equation}$$