1.2 Features of projectile motion  (Page 6/7)

Problem : A projectile, thrown at an angle 45° from the horizontal, strikes a building 30 m away at a point 15 above the ground. Find the velocity of projection.

Solution : As explained, the equation of projectile path suits the description of motion best. Here,

x = 30 m, y = 15 m and θ = 45°. Now,

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

$$\begin{array}{l}\Rightarrow 15=30\tan {45}^0-\frac{10x{30}^2}{2u^2{\cos }^2{45}^0}\\ \Rightarrow 15=30x1-\frac{10x2x{30}^2}{2u^2}\\ \Rightarrow 15=30-\frac{9000}{u^2}\\ \Rightarrow u^2=600\\ \Rightarrow u=24.49m/s\end{array}$$

Problem : A person standing 50 m from a vertical pole wants to hit the target kept on top of the pole with a ball. If the height of the pole is 13 m and his projection speed is 10√g m/s, then what should be the angle of projection of the ball so that it strikes the target in minimum time?

Solution : Equation of projectile having square of “tan θ” is :

$$\Rightarrow y=x\tan \theta -\frac{g{x}^2}{2u^2}\left(1+{\tan }^2\theta \right)$$

Putting values,

$$\Rightarrow 13=50\tan \theta -\frac{10x{50}^2}{2x{\left(10\sqrt{g}\right)}^2}\left(1+\tan {}^2\theta \right)$$

$$\Rightarrow 25{\tan }^2\theta -100\tan \theta +51=0$$

$$\Rightarrow \tan \theta =17/5or3/5$$

Taking the smaller angle of projection to hit the target,

$$\Rightarrow \theta ={\tan }^{-1}\left(\frac{3}{5}\right)$$

Problem : Determine the angle of projection for which maximum height is equal to the range of the projectile.

Solution : We equate the expressions of maximum height and range (H = R) as :

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

$$\begin{array}{l}{u}^2{\sin }^2\theta =2u^2\sin 2\theta \\ \Rightarrow {\sin }^2\theta =2\sin 2\theta =4\sin \theta x\cos \theta \\ \Rightarrow \sin \theta =4\cos \theta \\ \Rightarrow \tan \theta =4\\ \Rightarrow \theta ={\tan }^{-1}\left(4\right)\end{array}$$