Shows that the path of horizontal projectile is a parabola and derive an expression for (i) Time of flight (ii) Horizontal range (iii) resultant relative and any instant (iv) speed of the projectile when it hits the ground?

Explanation

(i) Consider an object thrown with initial velocity u at an angle `theta` with the horizontal.
(ii) Since acceleration due to gravity acts vertically downwards, velocity along the horizontal x-direction `u_(x)` doesn't change throught the motion. Whereas velocity along the y-direction `u_(y)` is changed.

The path of the projectile:
(a) Motion along x-direction:
(i) The horizontal distance travelled by the projectile at a point P after a time t can be written as,
`s_(x)=u_(x)t+(1)/(2)a_(x)t^(2)`
(ii) Here, `s_(x)=x, u_(x)=u cos theta, and a_(x)=0`, Therefore,
`x=u cos theta.t`
`t=(x)/(u cos theta)" " ...(1)`
(b) Motion along y-direction:
(i) The downward distance travelled by the projectile at a point P after a time t can be written as,
`s_(y)=u_(y)t+(1)/(2) a_(y)t^(2)`
(ii) Here, `s_(y)=y, u_(y)=u sin theta, and a_(y)=-g`
Therefore, `y=u sin theta t-(1)/(2)g t^(2)`
(iii) Substituting equation (1), we get,
`y=u sin theta (x)/(u cos theta)- (1)/(2) g((x)/(u cos theta))^(2)`
`y=x tan theta-(1)/(2) "g" (x^(2))/(u^(2) cos^(2)theta)`
(iv) Thus, the path travelled by the projectile is an inverted parabola.