This projectile motion takes place when the initial velocity is not horizontal, but at some angle with the vertical, as shown in Figure.
(Oblique projectile)
Examples:
• Water ejected out of a hose pipe held obliquely.
• Cannot fired in a battle ground.
Consider an object thrown with initial velocity at an angle `theta` with the horizontal.
Then,
`vecu = u_(x) hatl+u_(y)hatj`
where `u_(x) = u cos theta` is the horizontal component and `u_(y) = u sin theta` the vertical component of velocity
Since the acceleration due to gravity is in the direction opposite to the direction of vertical component `u_(y)` , this component will gradually reduce to zero at the maximum height of the projectile. At this maximum height, the same gravitational force will push the projectile to move downward and fall to the ground. There is no acceleration along the x direction throughout the motion. So, the horizontal component of the velocity (`u_(x) = u cos theta`) remains the same till the object reaches the ground.
Hence after the time t, the velocity along horizontal motion `v_(x) = u_(x) + a_(x)t = u_(x) = u cos theta`.
The horizontal distance travelled by projectile in time t is `s_(x) = u_(x)t + 1/2 a_(x)t^(2)` .
Here, `s_(x) = x, u_(x) = u cos theta, a_(x) = 0`
Thus, x = u cos `theta` t or t= `(x)/(u cos theta)` ...(1)
Next, for the vertical motion `y_(y) = u_(y) + a_(y)t`
Here `u_(y) = u sin theta, a_(y) = -g` (acceleration due to gravity acts opposite to the motion).
Thus, `v_(y), = u sin theta - "gt"`
The vertical distance travelled by the projectile in the same time t is
Here, `s_(y) = y, u_(y) = u sin theta, a_(x) = -g` Then,
`y = usin theta t 1/2 "gt"^(2)` ...(2)
Substitute the value of t from equation (i) in equation (ii), we have
`y = u sin theta (x)/(u cos theta) - 1/2 "g" (x^(2))/(u^(2) cos^(2) theta)`
`y = x tan theta - 1/2 "g" (x^(2))/(u^(2) cos^(2) theta)` ...(3)
Thus the path followed by the projectile is an inverted parabola.
Maximum height (`h_("max")`): The maximum vertical distance travelled by the projectile during the journey is called maximum height. This is determined as follows:
For the vertical part of the motion
`v_(y)^(2) = u_(y)^(2) + 2a_(y)s`
Here, `u_(y) = u sin theta, a= -g, s = h_("max")`, and at the maximum height `v_(y)` = 0
Hence, `(0)^(2) = u^(2) sin^(2)theta = 2 gh_("max")`
or `h_("max") = (u^(2) sin^(2) theta)/(2g)` ...(4)
Time of flight `(T_(f))`: The total time taken by the projectile from the point of projection till it hits the horizontal plane is called time of flight.
This time of flight is the time taken by the projectile to go from point O to B via point A as shown in figure.
We know that `s_(y) = u_(y)t+ 1/2 a_(y)t^(2)`
Here, `s_(s) = y = 0` (net displacement in y-direction is zero), `u_(y) = u sin theta, a_(y) = -g, t= T_(f)` Then
`0 = u sin thetaT_(f) - 1/2 gT_(f)^(2)`
`T_(f) = 2u (sin theta)/(g)` ...(5)
Horizontal range (R): The maximum horizontal distance between the point of projection and the point on the horizontal plane where the projectile hits the ground is called horizontal range (R). This is found easily since the horizontal component of initial velocity remains the same. We can write
Range R = Horizontal component of velocity x time of flight = `u cos theta xx T_(f) = vecr_(1) xx vecr_(2)`
The horizontal range directly depends on the initial speed (u) and the sine of angle of projection `(theta)`. It inversely depends on acceleration due to gravity .g..
For a given initial speed u, the maximum possible range is reached when sin `2theta` is maximum, `sin 2theta = 1`. This implies `2theta = pi//2`
or `theta (pi)/(4)`
This means that if the particle is projected at 45 degrees with respect to horizontal, it attains maximum range is given by.
`R_("max") = (u^(2))/(g)` ...(6)
