Projectile Motion MCQ - Practice Questions with Answers

• Projectile Motion-An object which is thrown with initial velocity & then it moves due to the effect of gravity alone is called projectile , it motion is called as projectile motion and its path called trajectory.

                E.g-  A javelin thrown by an athlete 

1. Projectile Projected at angle ϴ

• Initial Velocity-  U

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

• Final velocity = V 

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

So,

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

• Displacement = S

Horizontal component $=S_x=U \cos \theta \cdot t$
Vertical component $=S_y=U \sin \theta \cdot t-\frac{1}{2} \cdot g \cdot t^2$
and,

$$
S=\sqrt{S_x^2+S_y^2}
$$
 

•  Acceleration = a

Horizontal component = 0

Vertical component = -g

So, a = -g

 

Parameters in Projectile motion - 

1. Maximum Height - 

 

• Maximum vertical distance attained by a projectile during its journey.

• Formula-H=\frac{U^2 \sin ^2 \Theta}{2 g}

• When the velocity of the projectile increases n time then the Maximum height is increased by a factor of  $n^2$

• Special Case-

                                If U is doubled, H becomes four times provided & g is constant.

 

     2)  Time of Flight

• Time for which projectile remains in the air above the horizontal plane.

• Formula-

1.  $T=\frac{2 U \sin \theta}{g}$

2. Time of ascent =

                              $t_a=\frac{T}{2}$

 

3. Time of descent =

                                $t_d=\frac{T}{2}$

• When the velocity of the projectile increases n time then the Time of ascent becomes n times

• When the velocity of the projectile increases n time then the Time of descent becomes n times

• When the velocity of the projectile increases n time then the time of flight becomes n times.

 

3. Horizontal Range

• Horizontal distance traveled by projectile from the point of projectile to the point on the ground where it hits.

• Formula-

                         $R=\frac{u^2 \sin 2 \Theta}{g}$

• A special case of horizontal range

1. For max horizontal range.

    $\begin{aligned} \Theta & =45^0 \\ R_{\max } & =\frac{u^2 \sin 2(45)}{g}=\frac{u^2 \times 1}{g}=\frac{u^2}{g}\end{aligned}$

 

2. Range remains the same whether the projectile is thrown at an angle $\theta$ with the horizontal or at an angle $\theta$ vertical (90-$\theta$) with the horizontal 

3. When the velocity of the projectile increases n time then the horizontal range is increased by a factor of  $n^2$

4. When a horizontal range is n times the maximum height then 

                                                                                   $\tan \Theta=\frac{4}{n}$