Question

Range of projectile is R and maximum height covered by the particle is H, find area covered by the path of projectile and horizontal line​

Answer 1

Answer:

$\frac{2HR}{3}$

Explanation:

1) We have,

Equation of trajectory of path of projectile :

$y=xtan(\theta)[1-\frac{x}{R}]$

We have,

$H=\frac{u^2sin^2(\theta)}{2g} \\ \\R=\frac{u^2sin(2\theta)}{g}  \\ \\ => \frac{H}{R}=\frac{sin^2(\theta)}{2sin(2\theta)}\\ \\=> \frac{sin^2(\theta)}{2*2sin(\theta)cos(\theta)}=\frac{tan(\theta)}{4}\\ \\=>tan(\theta)=\frac{4H}{R}$

2) Now,

Area of Parabola is given by :

$\int\limits^R_0 {y}\,dx= \int\limits^R_0 {xtan(\theta)[1-\frac{x}{R}]} \, dx\\ \\=\frac{4H}{R}\int\limits^R_0 {x[1-\frac{x}{R}]} \, dx\\ \\=\frac{4H}{R} \int\limits^R_0 {(x-\frac{x^2}{R} )} \, dx =\frac{4H}{R} [\frac{R^2}{2}-\frac{R^3}{3R}]=\frac{2H}{3}$

Hence, The area covered by the path of projectile and horizontal line​ is given by 2H/3.