45 degrees below the ground II

Since the projectile is thrown horizontally, therefore the initial $\color{#20A900}{y-component}$ of its velocity is zero ;

Therefore at any height ' $\color{#3D99F6}{h}$ ' ,

$V_y$ $=$ $\sqrt {2gh}$ ...[from $3^{rd}$ $eq.^{n}$ of motion]

$\color{#3D99F6}{Now}$ ; At the instant of landing,

$\dfrac {V_y}{V_x}$ $=$ $\tan45^{\circ}$

$\Rightarrow$ $V_y$ $=$ $V_x$

$\Rightarrow$ $V_x$ $=$ $\sqrt {2gh}$

Plugging in $g$ $=$ $10$ \text {&} $h$ $=$ $20$ ;

we get $\rightarrow$

$\color{#69047E}{\boxed {V_x = 20}}$

At the landing point, the magnitud of the vertical and horizontal components of the velocity should be the same since the velocity vector makes and angle of 45 degrees with respect to the ground. The initial velocity was purely horizontal and since that component of the velocity remained unchanged through all the flight, the initial speed equals the magnitud of the vertical component of the projectile's velocity at the landing point.

The vertical position of the projectile, measured from the ground, is given by: $y=h- \frac{1}{2} g t^2$ Where "h" is the buildings height.

And its vertical component of the velocity is given by: $v_y=-gt$

Using the first equation for solving for the time of the projectile´s flight (y=0m at landing) and using the secong one for solving for the vertical component of the velocity and having in mind that: $\|v_0\|=\|v_y\|$ where $v_0$ is the projectile's initial velocity.

Finally, this yields: $\|v_0\|=\sqrt{2hg}$ and by replacing values for "h" and "g" we obtain that $\|v_0\|=20 \frac{m}{s}$ So, the answer is "20".

The projectile takes $\frac {2H}{g} = 2 s$ to reach the ground. Since the final velocity vector forms an angle of $45^{\circ}$ with the ground, $V_x = V_y$ . $\therefore V_x = 2g = 20$