Path of a projectile projected from the ground is given as y=ax-bx^2. find angle of projection, maximum height attained and horizontal range?
- RealProLv 75 years agoFavorite Answer
Soooo... under the assumption that ground is at y=0:
Find the derivative of the path function. This tells you the slope of the function at any x.
y' = a - (2b)x
y' = a
The angle that the function makes with the x-axis at that point (projection angle) is tan^-1(a)
For maximum height, slope = 0.
a - (2b)x = 0
x = a/(2b)
So maximum height = ax-bx^2 = a(a/(2b)) - b(a/(2b))^2 = a^2/(2b) - a^2/(4b)
Horizontal range is the non-zero value for x when y=0.
y=0 ===> ax-bx^2 = 0 ===> x(a - bx) = 0 ===> x = a/b = range