Projectile over a Square (part -1)

Since the horizontal velocity of a projectile is constant. Its horizontal displacement is directly proportional to time (\t) of travel.

Let center of the square be $O$ , the end of the projectory on the horizontal plane be $E$ , the time taken for the projectile to travel from $C$ to $D$ be $t_1$ and from $C$ to $E$ be \(t_2). Then, we have:

\(\dfrac {1}{2}gt_1^2= 10\sqrt{2}\quad \quad \dfrac {1}{2}gt_2^2= 20\sqrt{2}\quad \Rightarrow \dfrac {t_2^2}{t_1^2} = 2\quad \Rightarrow \dfrac {t_2}{t_1} = \sqrt{2}\)

$\Rightarrow \dfrac {AE}{OD} = \dfrac {t^2}{t_1} = \sqrt{2}\quad \Rightarrow AE = \sqrt{2}OD = 20 \space m$

Therefore, $\space R = 2\times AE = \boxed{40}\space m$ .

This problem is more geometry than mechanics. Since the path of the projectile is a parabola its equation taking the origin A midpoint of R due symmetry will be of the form

$x^{2}=a(d-y)$

we must determine the constants a and d.

For x= 0 ; y= d ; being d the diagonal of the square and for x=d/2 ; y=d/2 we get constant a=d/2.

Then we get
$x^{2}= d/2(d-y)$ so R values two times x when y=0 .

Then $R=d\sqrt{2 }$ ... and since $d= 20\sqrt{2}$ ... we get $R=40$ meters

It is easy to tell that the equation of the trajectory is $-10\sqrt { 2 } (y-20\sqrt { 2 } )={ x }^{ 2 }$ and since we want to find the range, $y=0$

Making $y$ the equation of the trajectory gives:

$-\frac { { x }^{ 2 }-400 }{ 10\sqrt { 2 } } =y$

Solving $-\frac { { x }^{ 2 }-400 }{ 10\sqrt { 2 } } =0$ gives

$x=20\quad or\quad -20$

Therefore, the answer is $20-(-20)=\boxed { 40 }$