202104160620 Homework 1 (Q4)

A particle is projected from a point $O$ on the horizontal floor. The range of the projectile is $R$ and the maximum height that the particle can reach is $h$ . Show that the equation of trajectory of the particle is

$\displaystyle{\frac{y}{h}=\frac{4x}{R}\bigg( 1-\frac{x}{R}\bigg)}$ .

Solution.

(The solution below is based on the manuscript of 2014-2015 PHYS1250 Fundamental Physics Homework Solutions.)

The trajectory of projectile motion must be a parabola, which can be expressed in the form of a quadratic equation:

$y=ax^2+bx+c$ ;

And since the particle passes through the points $(0,0)$ and $(R,0)$ , the equation of trajectory can be expressed in the form:

$y=A(x-0)(x-R)$ .

When the particle has traveled a horizontal distance $x=\displaystyle{\frac{R}{2}}$ , it reaches the maximum height $y=h$ .

Inserting the point $(\frac{R}{2},h)$ into the trajectory equation, we solve for the unknown $A$ :

$\begin{aligned} h & = A\bigg(\frac{R}{2}-0\bigg)\bigg(\frac{R}{2}-R\bigg) \\ h & = -\frac{AR^2}{4} \\ \Rightarrow \qquad A & = -\frac{4h}{R^2} \end{aligned}$

Thus,

$y = -\displaystyle{\frac{4h}{R^2}}x(x-R)$ ,

or,

$\boxed{\frac{y}{h} = \frac{4x}{R}\bigg( 1-\frac{x}{R}\bigg)}$

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