Horizontal Projectile Motion - Practice Questions & MCQ

1. Important equations

• Initial Velocity- u

Horizontal component $=u_x=u$
Vertical component $=u_y=0$
- Velocity ' $v$ ' after time ' $t$ ' sec-

Horizontal component $=v_x=u$
Vertical component $=v_y=g \cdot t$
and, $\quad v=\sqrt{v_x^2+v_y^2}$
i.e; $\quad v=\sqrt{u^2+(g t)^2}$
$\tan \beta=\frac{g t}{u}$
Where, $\beta=$ angle that velocity makes with horizontal
- Displacement=S

Horizontal component $=S_x=u . t$
Vertical component $=S_y=\frac{1}{2} g \cdot t^2$

and, $S=\sqrt{S_x^2+S_y^2}$
- Acceleration $=\mathrm{a}$

Horizontal component $=0$
Vertical component $=\mathrm{g}$
So, $a=g$

• Equation of path of a projectile

$
y=\frac{g}{2 u^2} \cdot x^2
$

$g \rightarrow \quad$ Acceleration due to gravity
$u \rightarrow$ initial velocity

2. Important Terms

1. Time of flight

• Formula

$
t=\sqrt{\frac{2 h}{g}}
$

where $t=$ time of flight
$h=$ Height from which projectile is projected
2. Range of projectile
- Formula

$
R=u \cdot \sqrt{\frac{2 h}{g}}
$

Where $R=$ Range of projectile
$u=$ horizontal velocity of projection from height h
3. Velocity at which projectile hit the ground.

$
v=\sqrt{u^2+2 g h}
$

Where $v=$ velocity at which projectile hit the ground.