Understanding Projectile Motion with Derivations

Projectile Motion is one of the most common concepts we encounter in our daily lives. Studying it becomes important because the ‘understanding’ (not mugging up formulae) that we take here comes in handy while analyzing complex concepts and experiments.

“Formulae give you Marks, but derivations give you understanding!

Another Fun article to read to understand the above quote would be the following: What is a Machine? -3 idiots machine definition

We already have a video on our Channel about such an experiment, on our channel, where knowledge of Projectile is needed for the Analysis part :

• Calculating e/m Ratio – Watch the full video on YouTube by clicking here

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1. Equation of Trajectory for the Projectile

We all have played ‘catch-catch,’ and we know what the motion of a ball looks like when it’s thrown. But how do you describe that curve mathematically? What’s the equation of that curve?

Most Important Tip that fixes everything related to Projectile!

Divide this whole 2-D situation into separate 1-D problems (X and Y)

• Please make sure to note down all quantities in X separately. This will be your separate problem
• Note down all quantities in Y separately. This will be your other separate problem
• At last, combine them to get results for 2-D motion.

Following the tip:

Starting from the origin (point where the ball is thrown) till point P (x,y):

Solving in the X direction:

$S_x = u_x t + \frac{1}{2} a_x t^2$

Substitute the values as discussed above

$x = u \cos\theta \times t$ –(1)

Solving in the Y direction:

$S_y = u_y t + \frac{1}{2} a_y t^2$

$y = u \sin\theta \cdot t + \frac{1}{2} (-g) t^2$ –(2)

Substitute values of ‘t’ from equation (1) into equation (2). This gives us an equation only in terms of x and y. This is what we refer to as ‘Locus’ or Equation of Trajectory here.

$y = x \tan\theta – \frac{1}{2} \frac{g x^2}{u^2 \cos^2\theta}$

The above equation is quadratic in nature. The Projectile Trajectory is a Parabola!

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2. Finding Expression for Horizontal Range

Range is basically the ‘horizontal’ distance that the ball covers (from the origin to the point where it lands). It means that we need to find R in the figure

• This can be easily found since x = 0 and x = R are the two roots of the parabola.
• To calculate the roots, simply put y = 0 in our Equation of Trajectory

Calculations :

Substituting y = 0 and taking ‘x’ common on the RHS of the Equation of Trajectory, we get the equation as:

$0 = x \left( \tan\theta – \frac{g x^2}{2u^2 \cos^2\theta} \right)$

Hence, we get some possibilities:

1st possibility:

$x=0$

This is quite obvious since the ball was at ground level before it was just thrown

2nd possibility:

$0 = \left( \tan\theta – \frac{g x^2}{2u^2 \cos^2\theta} \right)$

On Simplifying,

$x =R = \frac{u^2 \sin 2\theta}{g}$

* In the simplifying process, you need to use the $\sin 2\theta$ trigonometric identity

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3. Finding Time of Flight Expression

Let’s find out how much time the body stays in the projectile motion.

Note that :

• On landing, after completing the motion, the displacement in the Y direction is zero (Pause and observe !), since it again came to the same Y-level (S_y = 0 here)
• The time taken for this displacement in Y to become 0 is nothing but the Time of Flight (T)

In Y direction:

$S_y = u_y t + \frac{1}{2} a_y t^2$

As discussed, make the following substitutions: S_y = 0; t = T

$0 = u \sin\theta \cdot T + \frac{1}{2}(-g)T^2$

On simplifying,

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

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4. Finding Maximum Height Covered

If we look at the Equation of Continuity, in this case (simple throwing), the trajectory resembles a downward parabola since a<0

$y = x \tan\theta – \frac{1}{2} \frac{g x^2}{u^2 \cos^2\theta}$

Comparing this with $y = ax^2 + bx + c$, we would get,

$a = -\frac{g}{2u^2 \cos^2\theta}$

$b = \tan\theta$

$c = 0$

Hence, visualizing it graphically,

Finishing the Calculations,

• On calculating (-b/2a), we get x = R/2
  
• On calculating (-D/4a), we get:

$y = H = \frac{u^2 \sin^2\theta}{2g}$

So, this is how we get the expression for the Maximum Height reached, denoted by ‘H.’

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Conclusion

So, this is how we derive the expressions for Time of Flight, Range, Maximum Height, and Equation of Trajectory. Note that – With this same approach, we can solve almost any kind of problem related to Projectile Motion. And, why is that the case?

Because whatever we discussed in this blog/article is not a trick or something, it’s a complete concept with a complete understanding. This makes us equipped to solve any kind of problem (Throwing a ball from a cliff, a projectile on an incline, or just anything for that matter). You are now in a very good position to solve problems