Question

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?

Answer 1

See the explanation below

Explanation:

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)`