# Question #9243f

Sep 7, 2017

Consider it being thrown with an angle $\theta$ with the horizontal.

Then the velocity has two components ${u}_{x} = U \cos \theta$ and ${u}_{y} = U S \in \theta$.

The horizontal acceleration is ${a}_{x} = 0$ while that in vertical direction is ${a}_{y} = - g$

From an equation from kinematics,

$y = {u}_{y} t + \frac{1}{2} {a}_{y} {t}^{2}$
$\implies y = U S \in \theta t - \frac{1}{2} g {t}^{2}$ is distance travelled by particle in y direction (vertical direction) in time $t$.

Also, $x = {u}_{x} t + \frac{1}{2} {a}_{x} {t}^{2}$
$\implies x = U C o s \theta t$

Combining $x$ and $y$,

$y = x \tan \theta - \frac{1}{2} \frac{g {x}^{2}}{C} o {s}^{2} \theta$

But for $\theta$ constant, $\tan \theta = a$ and $b = - \frac{g}{2 C o {s}^{2} \theta}$,

We get,
$y = a x + b {x}^{2}$

Which shows that the path is parabolic.

The equations of motion are,

${F}_{x} = m {a}_{x} = 0$
${F}_{y} = m {a}_{y} = - m g$

If however, the body is thrown vertically up, $\theta = \frac{\pi}{2}$

And thats when the $x$ component of motion ceases the exist and, motion can be described as, by acceleration ${a}_{y} = - g$.

Then, $y = U t - \frac{1}{2} g {t}^{2}$ describes the vertical distance travelled.