Horizontal Projectile Motion MCQ - Practice Questions with Answers

1. Important equations

 

• Initial Velocity- U

       $\begin{aligned} & \text { Horizontal component }=U_x=U \\ & \text { Vertical component }=U_y=0\end{aligned}$

• Final velocity = V

        $\begin{aligned} & \text { Horizontal component }=V_x=U \\ & \text { Vertical component }=V_y=-g \cdot t\end{aligned}$

 

and

$$
V=\sqrt{V_x^2+V_y^2}
$$

i.e; $\quad V=\sqrt{U^2+g \cdot t^2}$
and,

$$
\tan \beta=\frac{g t}{U}
$$

Where $\beta=$ angle that velocity makes with horizontal

 

• Displacement=S      

Horizontal component $=S_x=u . t$
Vertical component $=$
and,

$$
S=\sqrt{S_x^2+S_y^2}
$$
 

• Acceleration = a

           Horizontal component= 0

           Vertical component = -g

           So, a = -g

 

• Equation of path of a projectile   

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

$g \rightarrow$ 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
$u=$ the horizontal velocity of projection from height $h$

 

3. Velocity at which projectile hit the ground.

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

           Where velocity at which projectile hit the ground.