Projectile motion

The vertical displacement $y$ of the projectile at time $t$ is given by:

$\begin{aligned} y & = u_y t- \frac 12 gt^2 \quad \quad \small \color{#3D99F6} \text{where }u_y \text{ is the vertical component of initial velocity.} \\ & = 20\sin 45^\circ t- \frac 12 (10)t^2 \\ \implies y & = 10\sqrt 2t- 5t^2 \quad …(1) \end{aligned}$

The horizontal displacement $x$ :

$\begin{aligned} x & = u_x t \\ &= 20\cos 45^\circ \\ &= 10\sqrt 2t \\ \implies t &= \frac x {10\sqrt 2} \quad …(2) \end{aligned}$

Substituting (2) in (1):

$\begin{aligned} \implies y & = 10\sqrt 2t- 5t^2 \\ &= x -\frac {x^2}{40} &\small \color{#3D99F6} \text{Putting }x=10 \\ &=10-\frac {10^2}{40} \\ &= \boxed{7.5} \end{aligned}$

First analyze the projectile horizontally: find the time $t$ when the ball hits the wall $10 \ m$ away.

$\displaystyle \begin{aligned} d_x & = v_xt \\ 10 & = 20\cos45^{\circ}t \\ t & = \frac{1}{\sqrt{2}} \end{aligned}$

Now analyze the projectile vertically: find the vertical distance $d_y$ where the ball hits the wall at time $\displaystyle t = \frac{1}{\sqrt{2}}$ .

$\displaystyle \begin{aligned} d_y & = v_it + \frac12at^2 \\ & = 20\sin45^{\circ}\left(\frac{1}{\sqrt{2}}\right) + \frac12(- 10)\left(\frac{1}{\sqrt{2}}\right)^2 \\ & = 10 - \frac52 \\ & = \boxed{7.5 \ m} \end{aligned}$

The trajectory of the ball is given by the equation:
$y=\tan{\theta}\cdot x - \frac{g}{2v^{2}\cdot (\cos{\theta})^{2}}\cdot x^{2}$
where $\theta$ is the angle that we throw the ball, $g$ the acceleration of gravity, $v$ the speed at which we throw the ball, $y$ the height and $x$ the horizontal distance.
If we substitute:
$\theta=45^{\circ}$ , $x=10m$ , $g=10\frac{m}{s^{2}}$ , $v=20\frac{m}{s}$ we find that:
$y=7.5m$
Therefore the ball will hit the wall at a height of 7.5 meters

Proof of the trajectory equation
The initial speed on the x axis is : $v_{x}=v\cdot \cos{\theta}$
So: $x=v_{x}\cdot t\Rightarrow x=v\cdot \cos{\theta}\cdot t\Rightarrow t=\frac{x}{v\cdot \cos{\theta}}$
The inital speed on the y axis is: $v_{y}=v\cdot \sin{\theta}$
So: $y=v_{y}\cdot t - \frac{1}{2}\cdot g \cdot t^{2}\Rightarrow$
$y=v\cdot \sin{\theta}\cdot \frac{x}{v\cdot \cos{\theta}} - \frac{1}{2}\cdot g \cdot (\frac{x}{v\cdot \cos{\theta}})^{2}\Rightarrow$
$y=\tan{\theta}\cdot x - \frac{g}{2\cdot v^{2}\cdot (\cos{\theta})^{2}}\cdot x^{2}$