Everytime I approach any projectile motion/kinematics problem, I get confused. I don't know how to translate the problem into an operational method, and every time I complete a problem, the next one is a new mystery to me.

How should I tackle this issue and master this problem type?

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    $\begingroup$ Assuming that air drag is ignored, there is an absolutely KEY concept that makes such problems easier to investigate. Whatever happens in the vertical direction is totally independent of whatever happens in the horizontal direction. This means that you can break such problems into two parts, whereby you apply constant motion kinematic equations in the horizontal direction and accelerated motion kinematic equations in the vertical direction. $\endgroup$ Oct 8, 2023 at 23:49

2 Answers 2


All projectile motion questions are the same.

Why do the kinematics equations make sense?

We have two key equations in kinematics, and they stem from the mathematical relationship between position, momentum, and velocity.

Velocity is defined as $v = \frac{dx}{dt}$. Acceleration is defined as $a = \frac{dv}{dt}$. Importantly, the derivative of a function introduces some redundancy. Consider the equation $v(t) = 2t$, then $$\int v(t)dt = \int 2t dt = t^2 + c.$$ Now, if we know that a particle was initially at rest, we know that $v(t = 0) = 0$. We can use this to find c! $$v(0) = 0^2 + c = 0 \rightarrow c = 0$$

Thus, our initial position and velocity are actually integration constants. This physically explains $$v(t) = v_0 + at$$ $$x(t) = x_0 + v_0t + \frac{1}{2} at^2$$ These are the only unique pieces of information. You can plug these in in crafty ways to solve the problem at hand.

Operational Approach: How do I solve these problems systematically?

With that under our belt, we can ask “how do I solve my projectile motion problem?”. The following approach will always work for ANY projectile motion problem.

  1. Identify what the question is asking for.
  2. Identity, mathematically, what you know. Is velocity initially at rest? Then $v(0) = 0m/s$. Is your particle moving only in the vertical direction? Then $a_x = 0m/s^2$. Is acceleration due to gravity? Then $a_y = g$. Is the height of the cliff 5m? Then $y(0) = 5m$. And the list goes on.
  3. Write down the kinematics equations (if you don’t want to memorize them, rederive them according to the logic above. Otherwise, just remember them).
  4. You should now have a list of quantities and equations. Your task is purely to rearrange the equations you have written in (3) to solve for the quantities you identified in (1). If you have done 1-3 correctly, you will be able to achieve this goal.

Amplifying on Step 3 of @Relativisticcucumber excellent answer:

There are only five variables in any constant acceleration one-dimensional motion problem:

  1. Distance
  2. Acceleration
  3. Final velocity
  4. Time
  5. Initial velocity

Various sets of symbols are widely used for these variables; many years ago I learned "s a v t u", possibly because that's how my Physics teacher learned them, because that's how his...

Similarly there are five distinct equations that involve these five variables: https://en.wikipedia.org/wiki/Equations_of_motion

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Significantly, each of these equations uses only four of the five variables; each of the variables is missing from one of the equations.

In approaching any problem, then, you need to identify:

  1. The three variables whose values you are given and those values;
  2. The variable whose value you are asked to find;
  3. The variable that you neither know nor want to know

If you are asked to find both of the missing variables, then you are being asked two questions. Pick one of the unknowns to solve for, do it, then do the other.

Another necessary step in solving this type of problem is the establishment of a sign convention. The first thing you should write down as you set up a solution is "Up is positive!", or "Motion to the left is positive!"

It is purely an arbitrary choice; you could just as easily decide "Up is negative!". But you must use the same choice for each variable in the problem, and you must interpret any answer in the same light.

For example:

"You throw a stone upward from the edge of a cliff at $30 \text{ m/s}$. Where will the stone be after $12 \text{ s}$? (g = $10 \text{ }m/s^2$)

The "unused" variable is $v$, final velocity. So using Equation 2.

If "Up is positive!" then u = $30$, g = $-10$, t = $12$ and thus s = $-360$. Since up is positive, the stone is 360 m below the edge of the cliff.


If "Down is positive, then then u = $-30$, g = $10$, t = $12$ and thus s = $360$. Since down is positive, the stone is 360 m below the edge of the cliff

  • $\begingroup$ Another common mnemonic is "suvat". $\endgroup$
    – PM 2Ring
    Oct 9, 2023 at 5:44

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