A projectile is thrown horizontally from a $20\text{ m}$ building. If, at the landing point, the velocity vector of the projectile makes an angle of $45^\circ$ below the ground, calculate the projectile's initial speed in $\text{m/s}$ .

Take $g$ (the accelaration due to gravity) as $10 \text{ m/s}^2$ .

The answer is 20.

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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 $\rightarrow$

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