Question

If y = x-x² is the equation of trajectory of a projectile , the horizontal range of the projectile will be

Answer 1

Explanation:

The function is

y=ax−bx2

ANGLE OF PROJECTION

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

dydx=a−2bx

x=0, ⇒, dydx(0)=a

So, the angle of projection is arctan(a)

HORIZONTAL RANGE

The horizontal range is when y=0

ax−bx2=0

x(a−bx)=0

Therefore,

x=0 which is the initial point

and x=ab

The range is =ab

MAXIMUM HEIGHT

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

x=a2b

Therefore,

y=a⋅a2b−b⋅(a2b)2=

Answer 2

y=0 in the point where it falls to the ground.

the distance between initial point and this is the horizontal range.so, 0=x-x^2

x(x-1)=0

either     x=0 or x=1

x=0 is the initial position where it starts from the ground.

therefore the range is 1 unit.