Show that the trajectory of an object thrown at certain angle with the horizontal is a parabola .

Explanation

Projectile : A body thrown into the air same angle with the horizontal, (other tan `90^(@)` ) its motion under the influence of gravity is called
projectile . The path followed by it is called
trajectory. Let a body is projected from point O, with velocity 'u' at an angle `theta` with horizon-tal . The velocity 'u' can be resolved into two rectangular components `u_(x)` and `u_(y)` along x-axis and y-axis .
`u_(x) = u cos theta` and `u_(y) = u sin theta `
After time t, Horizontal distance travelled
` x = cos theta` . t ........(1)
After a time 't' sec, vertical distacement
y = u sin `theta* t - (1)/(2) gt^(2)`
From (1) t = `(x)/(u c0s theta)`
`y = (u sin theta)/(u cos theta) - (1)/(2)g* (x^(2))/(u^(2) cos ^(2) theta)`
`rArr y = tan theta - (gx^(2))/(2u^(2)cos^(2)theta)`
Let tan `theta` = A and `(g)/(2u^(2)cos^(2)theta) = B ` then
`y = Ax - Bx^(2)`
The above equation represents "parabola" . Hence the path of a projectile is a parabola .