1.2 Features of projectile motion  (Page 5/7)

Problem : A projectile is thrown with a velocity of $25\sqrt{2}$ m/s and at an angle 45° with the horizontal. The projectile just clears two posts of height 30 m each. Find (i) the position of throw on the ground from the posts and (ii) separation between the posts.

Solution : Here, we first use the equation of displacement for the given height in the vertical direction to find the values of time when projectile reaches the specified height. The equation of displacement in vertical direction (y) under constant acceleration is a quadratic equation in time (t). Its solution yields two values for time. Once two time instants are known, we apply the equation of motion for uniform motion in horizontal direction to determine the horizontal distances as required. Here,

$$\begin{array}{l}{u}_y=25\sqrt{2}X\sin {45}^0=25m/s\end{array}$$

$$\begin{array}{l}y=u_{y}t-\frac{1}{2}g{t}^2\\ \Rightarrow \mathrm{30}=\mathrm{25}t-\frac{1}{2}X\mathrm{10}X{t}^2\\ \Rightarrow t^2-5t+6=0\\ \Rightarrow t=2s\mathrm{or}3s\end{array}$$

Horizontal motion (refer the figure) :

$$\begin{array}{l}\mathrm{OA}=u_{x}t==25\sqrt{2}x\cos {45}^{0}X2=50m\end{array}$$

Thus, projectile needs to be thrown from a position 50 m from the pole. Now,

$$\begin{array}{l}\mathrm{OB}=u_{x}t==25\sqrt{2}x\cos {45}^{0}X3=75m\end{array}$$

Hence, separation, d, is :

$$\begin{array}{l}d=\mathrm{OB}-\mathrm{OA}=75-50=25m\end{array}$$

Alternatively

Equation of vertical displacement (y) is a quadratic equation in horizontal displacement (x). Solution of equation yields two values of "x" corresponding to two positions having same elevation. Now, equation of projectile is given by :

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

Putting values,

$$\begin{array}{l}\mathrm{30}=x\tan \mathrm{45°}-\frac{10x^2}{2{\sqrt{\mathrm{25}}}^2{\cos }^2\mathrm{45°}}\\ \Rightarrow \mathrm{30}=x-\frac{10x^2}{2{25\sqrt{2}}^2{\cos }^2\mathrm{45°}}\\ \Rightarrow \mathrm{30}=x-\frac{x^2}{125}\\ \Rightarrow x^2-\mathrm{125}x+\mathrm{3750}=0\\ \Rightarrow (x-\mathrm{50})(x-\mathrm{75})=0\\ \Rightarrow x=\mathrm{50}\mathrm{or}\mathrm{75}m\\ \Rightarrow d=\mathrm{75}-\mathrm{50}=\mathrm{25}m\end{array}$$