Derive expression for maximum height, time of flight and range of a projectile.

Explanation

The maximum vertical distance travelled by the projectile during its journey is called maximum height.
(ii) For the vertical part of the motion,
`v_(y)^(2)=u_(y)^(2)+2a_(y)s_(y)`
(iii) Here, `V_(y)=0, S=_(y)=h_("max"), u_(y)=u and a_(y)=-g`
Therefore.
`0=u^(2)sin^(2)theta-2gh_("max")," "h_("max")=(u^(2)sin^(2)theta)/(2g)`
Time of flight : `(T_(f))`
(i) The time of flight `(T_(f))` is the time taken by the projectile to hit the ground after thrown.
(ii) The downward distance travelled by the projectile at a time t can be written as,
`s_(y)=u_(y)t+(1)/(2)a_(y)t^(2)`
(iii) Here substituting the values `S_(y)=0, t=T_(f).u_(y)=u sin theta, and a_(y)=-g ` we get,
`0=u sin theta-(1)/(2)gT_(f)^(2)`
Therefore, `T_(f)=(2u sin theta)/(g)`
Horizontal range : (R)
(i) The horizontal range (R) is the maximum horizontal distance distance between the point of projection and the point where the projectile hits the ground.
(ii) The horizontal distance travelled by the projectile at a time t can be written as,
`s_(x)=u_(x)t+(1)/(2)a_(x)t^(2)`
(iii) Here, `S_(x)=R, u_(x)=u cos theta, a_(x)=0 and t=T_(f)`
`R=u cos theta.T_(f)`
`R=u cos theta.(2u sintheta)/(g)=(2u^(2)sin theta cos theta)/(g)`
`" "[because T_(f)=(2usin theta)/(g)]`
(iv) Therefore, `" "F=(u^(2)sin 2theta)/(g)`
`" "[because sin 2theta=2 sin theta. cos theta]`
(v) For maximum range is, `R-(u^(2))/(g)`