5.3 Features of projectile motion (application)  (Page 2/3)

Problem : A bullet from a gun is fired at a muzzle speed of 50 m/s to hit a target 125 m away at the same horizontal level. At what angle from horizontal should the gun be aimed to hit the target (consider g = 10 $m/s^2$ ) ?

Solution : Here, horizontal range is given. We can find out the angle of projection from horizontal direction, using expression of horizontal range :

$$\begin{array}{l}R=\frac{u^2\sin 2\theta }{g}\\ \Rightarrow \sin 2\theta =\frac{gR}{u^2}=\frac{10x125}{{50}^2}=\frac{1}{2}\\ \Rightarrow \sin 2\theta =\sin {30}^0\\ \Rightarrow \theta ={15}^0\end{array}$$

Problem : A projectile has same horizontal range for a given projection speed for the angles of projections ${\theta }_1$ and ${\theta }_2({\theta }_2>{\theta }_1)$ with the horizontal. Find the ratio of the times of flight for the two projections.

Solution : The ratio of time of flight is :

$$\begin{array}{l}\frac{T_1}{T_2}=\frac{u\sin {\theta }_1}{u\sin {\theta }_2}=\frac{\sin {\theta }_1}{\sin {\theta }_2}\end{array}$$

For same horizontal range, we know that :

$$\begin{array}{l}{\theta }_2=({90}^0-{\theta }_1)\end{array}$$

Putting this, we have :

$$\begin{array}{l}\frac{T_1}{T_2}=\frac{\sin {\theta }_1}{\sin ({90}^0-{\theta }_1)}=\tan {\theta }_1\end{array}$$

Problem : A projectile, thrown at an angle 15° with the horizontal, covers a horizontal distance of 1000 m. Find the maximum distance the projectile can cover with the same speed (consider g = 10 $m/s^2$ ).

Solution : The range of the projectile is given by :

$$\begin{array}{l}R=\frac{u^2\sin 2\theta }{g}\end{array}$$

Here, R = 1000 m, g = 10 $m/s^2$ , θ = 15°

$$\begin{array}{l}\Rightarrow 1000=\frac{u^2\sin (2x{15}^0)}{g}=\frac{u^2}{2g}\\ \Rightarrow \frac{u^2}{2g}=1000m\end{array}$$

Now, the maximum range (for θ = 45°) is :

$$\begin{array}{l}\Rightarrow R_{\max }=\frac{u^2}{g}=2000m\end{array}$$

Problem : Two balls are projected from the same point in the direction inclined at 60° and 30° respectively with the horizontal. If they attain the same height, then the ratio of speeds of projection is :

Solution : Since the projectiles attain same height,

$$\begin{array}{l}{H}_1=H_2=H\\ \frac{{u_1}^2{\sin }^2{60}^0}{2g}=\frac{{u_2}^2{\sin }^2{30}^0}{2g}\\ \frac{{u_1}^{2}X3}{4}=\frac{{u_2}^{2}X1}{4}\\ {u_1}^2:{u_2}^2=1:3\\ \Rightarrow u_1:u_2=1:\sqrt{3}\end{array}$$

Problem : A projectile is thrown vertically up, whereas another projectile is thrown at an angle θ with the vertical. Both of the projectiles stay in the air for the same time (neglect air resistance). Find the ratio of maximum heights attained by two projectiles.

Solution : Let $u_1$ and $u_2$ be the speeds of projectiles for vertical and non-vertical projections. The times of the flight for vertical projectile is given by :

$$\begin{array}{l}{T}_1=\frac{2u_1}{g}\end{array}$$

We note here that the angle is given with respect to vertical - not with respect to horizontal as the usual case. As such, the expression of time of flight consists of cosine term :

$$\begin{array}{l}{T}_2=\frac{2u_2\cos \theta }{g}\end{array}$$

As, time of flight is same,

$$\begin{array}{l}\frac{2u_1}{g}=\frac{2u_2\cos \theta }{g}\\ u_1=u_2\cos \theta \end{array}$$

On the other hand, the maximum heights attained in the two cases are :

$$\begin{array}{l}{H}_1=\frac{{u_1}^2}{2g}\\ H_2=\frac{{u_2}^2{\cos }^2\theta }{2g}\end{array}$$

Using the relation $u_1=u_2\cos \theta$ as obtained earlier, we have :

$$\begin{array}{l}{H}_2=\frac{{u_1}^2}{2g}\\ \Rightarrow H_1=H_2\end{array}$$

Problem : The times for attaining a particular vertical elevation during projectile motion are $t_1$ and $t_2$ . Find time of flight, T, in terms of $t_1$ and $t_2$ .

Solution : We can answer this question analytically without using formula. Let the positions of the projectile at two time instants be "A" and "B", as shown in the figure. The time periods $t_1$ and $t_2$ denotes time taken by the projectile to reach points "A" and "B" respectively. Clearly, time of flight, T, is equal to time taken to travel the curve OAB ( $t_2$ ) plus the time taken to travel the curve BC.

Now, projectile takes as much time to travel the curve OA, as it takes to travel curve BC. This is so, because the time of travel of equal vertical displacement in either direction (up or down) in vertical motion under gravity is same. Since the time of travel for curve OA is $t_1$ , the time of travel for curve BC is also $t_1$ . Thus, the total time of flight is :

$$\begin{array}{l}T=t_1+t_2\end{array}$$

Alternatively,

As the heights attained are equal,

$$\begin{array}{l}{h}_1=h_2\\ \Rightarrow u_y{t}_1-\frac{1}{2}g{t_1}^2=u_y{t}_2-\frac{1}{2}g{t_2}^2\\ \Rightarrow u_y(t_2-t_1)=\frac{1}{2}g({t_2}^2-{t_1}^2)=\frac{1}{2}g(t_2+t_1)(t_2-t_1)\\ \Rightarrow t_2+t_1=\frac{2u_y}{g}=T\end{array}$$