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

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

• Final velocity = V 

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

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.  $T=\frac{2U\sin \theta }{g}$

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-

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

• A special case of horizontal range

1. For max horizontal range.

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

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 

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