Path of a projectile projected from the ground is given as y = ax-bx^2. What is the angle of projection, maximum height attained and horizontal range of the projectile?

Question

17920 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-15T13:40:30

See the explanation below

The function is

$y=ax-bx^2$

ANGLE OF PROJECTION

The angle of projection is obtained by calculating the gradient when $x=0$

$dy/dx=a-2bx$

$x=0$ , $=>$ , $dy/dx(0)=a$

So, the angle of projection is $arctan(a)$

HORIZONTAL RANGE

The horizontal range is when $y=0$

$ax-bx^2=0$

$x(a-bx)=0$

Therefore,

$x=0$ which is the initial point

and $x=a/b$

The range is $=a/b$

MAXIMUM HEIGHT

The x-coordinate of the maximum point is half the range.

$x=a/(2b)$

Therefore,

$y=a*a/(2b)-b*(a/(2b))^2=a^2/(2b)-a^2/(4b)=a^2/(4b)$

The maximum height is $=a^2/(4b)$