How Far Did I Throw It?

Applying Newton's equation, $s = ut + \dfrac{1}{2}at^2$ in horizontal and vertical direction.
In horizontal Direction , $\large x = t\sqrt{2gh} + 0 \quad \quad \quad .. \boxed{1}$ In vertical direction , $\large h = \dfrac{1}{2}gt^2 \quad \quad \quad \quad ..\boxed{2}$ Substituting value of t from $\boxed{2}$ in $\boxed{1}$ ,
$\large x = \sqrt{\dfrac{2gh \times 2h}{g}}$ $\large \boxed{\color{#3D99F6}{x = 2h}}$

There is a very important point is that the projectile has thrown parallel to $\color{#D61F06}{X-axis}$ . So the vertical component $v_0 sin\theta = v_0 sin 0^\circ = 0$ ..Now, $h= v_0 sin \theta t +\frac{1}{2} g t^2$ $\Rightarrow h=\frac{1}{2} g t^2$ $\Rightarrow t= \sqrt{\frac{2h}{g}}$ Then the time came to find out the distance traveled on $\color{#D61F06}{X -axis}$ .. As this component is constant always, We can use the formula of uniform velocity. Then the equation forms, $x =v_0 t$ $\Rightarrow x= \sqrt{2gh * \frac{2h}{g}}$ $\Rightarrow \boxed{\color{#E81990}{x= 2h}}$

We know that,

• $x = vt~~~~ ...(1)$ where $x$ is the distance.

• $h = ut+ \dfrac 12 gt^2~~~~ ....(2)$ where $u = 0.$

We have $v = \sqrt{2gh},$ now from $(1):$

$x = vt \\ x = \sqrt{2gh} \cdot t \implies \dfrac x{\sqrt{2gh}} = t.$

Substituting the value of $t$ in $(2):$

$\begin{aligned} h & = \dfrac g2\left(\dfrac x{\sqrt{2gh}}\right)^2 \\ h& = \dfrac g2\left(\dfrac {x^2}{2gh}\right) \\ h & = \dfrac{gx^2}{4gh} \\ h& = \dfrac{x^2}{4h} \\ 4h^2 &= x^2\\ \implies x & = \boxed{2h} \\ \end{aligned}$