5.1 Projectile motion (application)

Problem : A projectile is thrown with a speed of 15 m/s making an angle 60° with horizontal. Find the acute angle, "α", that it makes with the vertical at the time of its return on the ground (consider g = 10 $m/s^2$ ).

Solution : The vertical component of velocity of the projectile at the return on the ground is equal in magnitude, but opposite in direction. On the other hand, horizontal component of velocity remains unaltered. The figure, here, shows the acute angle that the velocity vector makes with vertical.

The trajectory is symmetric about the vertical line passing through point of maximum height. From the figure, the acute angle with vertical is :

$$\begin{array}{l}\alpha ={90}^0-\theta ={90}^0-{60}^0={30}^0\end{array}$$

Problem : Motion of a projectile is described in a coordinate system, where horizontal and vertical directions of the projectile correspond to x and y axes. The velocity of the projectile is 12 i + 20 j m/s at an elevation of 15 m from the point of projection. Find the maximum height attained by the projectile (consider g = 10 $m/s^2$ ).

Solution : Here, the vertical component of the velocity (20 m/s) is positive. It means that it is directed in positive y-direction and that the projectile is still ascending to reach the maximum height. The time to reach the maximum height is obtained using equation of motion in vertical direction :

$$v_y=u_y-gt$$

$$\Rightarrow 0=20-\mathrm{10}t$$ $$\Rightarrow t=2s$$

Now, the particle shall rise to a vertical displacement given by :

$$\begin{array}{l}{y}^{‘}=u_{y}t-\frac{1}{2}g{t}^2=20x2-5x{2}^2=20m\end{array}$$

The maximum height, as measured from the ground, is :

$$\begin{array}{l}H=15+20=35m\end{array}$$

Problem : The equation of a projectile is given as :

$$\begin{array}{l}y=\sqrt{3}x-\frac{1}{2}g{x}^2\end{array}$$

Then, find the speed of the projection.

Solution : The general equation of projectile is :

$$\begin{array}{l}y=x\tan \theta -\frac{g{x}^2}{2u^2{\cos }^2\theta }\end{array}$$

On the other hand, the given equation is :

$$\begin{array}{l}y=\sqrt{3}x-\frac{1}{2}g{x}^2\end{array}$$

Comparing two equations, we have :

$$\begin{array}{l}\tan \theta =\sqrt{3}\\ \Rightarrow \theta ={60}^0\end{array}$$

Also,

$$\begin{array}{l}{u}^2{\cos }^2\theta =1\\ \Rightarrow u^2=\frac{1}{{\cos }^2\theta }\\ \Rightarrow u^2=\frac{1}{{\cos }^{2}60}=4\\ \Rightarrow u=2m/s\end{array}$$

Problem : A projectile is projected at an angle 60° from the horizontal with a speed of ( $\sqrt{3}+1$ ) m/s. The time (in seconds) after which the inclination of the projectile with horizontal becomes 45° is :

Solution : Let "u" and "v" be the speed at the two specified angles. The initial components of velocities in horizontal and vertical directions are :

$$\begin{array}{l}{u}_x=u\cos {60}^0\\ u_y=u\sin {60}^0\end{array}$$

Similarly, the components of velocities, when projectile makes an angle 45 with horizontal, in horizontal and vertical directions are :

$$\begin{array}{l}{v}_x=v\cos {45}^0\\ v_y=v\sin {45}^0\end{array}$$

But, we know that horizontal component of velocity remains unaltered during motion. Hence,

$$\begin{array}{l}{v}_x=u_x\\ \Rightarrow v\cos {45}^0=u\cos {60}^0\\ \Rightarrow v=\frac{u\cos {60}^0}{\cos {45}^0}\end{array}$$

Here, we know initial and final velocities in vertical direction. We can apply v = u +at in vertical direction to know the time as required :

$$\begin{array}{l}v\sin {45}^0=u+at=u\sin {60}^0-gt\\ \Rightarrow v\cos {45}^0=u\cos {60}^0\\ \Rightarrow t=\frac{u\sin {60}^0-v\sin {45}^0}{g}\end{array}$$

Substituting value of "v" in the equation, we have :

$$\begin{array}{l}\Rightarrow t=\frac{u\sin {60}^0-u\left(\frac{\cos {60}^0}{\cos {45}^0}\right)X\sin {45}^0}{g}\\ \Rightarrow t=\frac{u}{g}(\sin {60}^0-\cos {60}^0)\\ \Rightarrow t=\frac{(\sqrt{3}+1)}{10}\left\{\frac{(\sqrt{3}-1)}{2}\right)\\ \Rightarrow t=\frac{2}{20}=0.1s\end{array}$$

Problem : A projectile is thrown with an angle θ from the horizontal with a kinetic energy of K Joule. Find the kinetic energy of the projectile (in Joule), when it reaches maximum height.

Solution : At the time of projection, the kinetic energy is given by :

$$\begin{array}{l}K=\frac{1}{2}m{u}^2\end{array}$$

At the maximum height, vertical component of the velocity is zero. On the other hand, horizontal component of the velocity of the particle does not change. Thus, the speed of the particle, at the maximum height, is equal to the magnitude of the horizontal component of velocity. Hence, speed of the projectile at maximum height is :

$$\begin{array}{l}v=u\cos \theta \end{array}$$

The kinetic energy at the maximum height, therefore, is :

$$\begin{array}{l}{K}^{‘}=\frac{1}{2}m{\left(u\cos \theta \right)}^2\end{array}$$

Substituting value of "u" from the expression of initial kinetic energy is :

$$\begin{array}{l}{K}^{‘}=\frac{mx2xK}{2m}{\cos }^2\theta \\ K^{‘}=K{\cos }^2\theta \end{array}$$

Problem : A projectile with a speed of “u” is thrown at an angle of “θ” with the horizontal. Find the speed (in m/s) of the projectile, when it is perpendicular to the direction of projection.

Solution : We need to visualize the direction of the projectile, when its direction is perpendicular to the direction of projection. Further, we may look to determine the direction of velocity in that situation.

The figure, here, shows the direction of velocity for the condition, when the direction of projectile is perpendicular to the direction of projection. From ΔOAB,

$$\begin{array}{l}\mathrm{\angle OBA}={180}^0-({90}^0+\theta )={90}^0-\theta \end{array}$$

Thus, the acute angle between projectile and horizontal direction is 90- θ for the given condition. Now, in order to determine the speed, we use the fact that horizontal component of velocity does not change.

$$\begin{array}{l}v\cos ({90}^0-\theta )=u\cos \theta \\ \Rightarrow v\sin \theta =u\cos \theta \\ \Rightarrow v=u\cot \theta \end{array}$$