`\color{red} ✍️` An object that is in flight after being thrown or projected is called a `text(projectile)`. Such a projectile might be a football, a cricket ball, a baseball or any other object.

`\color{red} ✍️` The motion of a projectile may be thought of as the result of two separate, simultaneously occurring components of motions.

• In our discussion, we shall assume that the air resistance has negligible effect on the motion of the projectile.

Suppose that the projectile is launched with velocity `vecv_o` that makes an angle `theta_o` with the x-axis as shown in Fig.

After the object has been projected, the acceleration acting on it is that due to gravity which is directed vertically downward:
`veca=-ghatj`
Or, `a_x=0, a_y=-g`

`color(green)("The components of initial velocity")` `vecv_o` are:
`color(green)(v_(o x)=v_o cos theta_o)`

`color(green)(v_(o y)=v_o sin theta_o)`

If we take the initial position to be the origin of the reference frame as shown in Fig. We have,
`x_o=0, y_o=0`

Then, `x=x_0+v_(0x)t+1/2a_xt^2`

`x=v_(o x)t=(v_o cos theta_o)t`

and `y=y_0+v_(0y)t+1/2a_yt^2`

`y=(v_o sin theta_o)t-1/2 g t^2`

These equations give the x-, and y-coordinates of the position of a projectile at time t in terms of two parameters — initial speed `v_o` and projection angle `theta_o`.

• One of the components of velocity, i.e. x-component remains constant throughout the motion and only the y- component changes, like an object in free fall in vertical direction.

The components of velocity at time t can be obtained using `v_x = v_(o x) + a_xt` and `v_y= v_(oy) a_yt`.
so, `v_x=v_(o x)=v_ocos theta_o`

`v_y=v_o sin theta_o -g t`

• Note that at the point of maximum height, `v_y=0`, `:. theta=tan^(-1)((v_y)/(v_x))=0`

As we discussed earlier, `x=v_(o x)t=(v_o cos theta_o)t` and `y=(v_o sin theta_o)t-1/2 g t^2`.

Now eliminating the time between the expressions for x and y, we obtain :

`color(blue)(y=(tantheta_o) x -g/(2(v_o cos theta_o)^2) x^2)`

Since `g,theta_o` and `v_o` are constants,

we can say that this equation is of the form `y=ax+by^2`, in which `a` and `b` are constants.

This equation represents to a `color(red)(text(parabola))`, i.e. `color(green)("the path of the projectile is a parabola.")`

Let `color(green)("time of maximum height"=t_m`

At the point of maximum height `v_y=0`

we have equation, `color(red)(v_y = v_o sin theta_o – g t)`

So, `v_y = v_o sin theta_o – g t_m=0`

`t_m=(v_o sin theta_o)/g`

`\color{fuchsia} \★mathbf\ul"Time of Flight"`

The total time `T_f` during which the projectile is in flight can be obtained by putting `y=0` in the equation

`y = (v_o sin theta_o ) t – 1/2 g t^2`.

`color(blue)(T_f= (2(v_o sin theta_o))/g)`

`color{green}{T_f" is known as the time of flight of the projectile."}`

• Note that `color{red}{T_f = 2 t_m}` (because of the symmetry of the parabolic path)

The maximum height `color(red)(h_m)` reached by the projectile can be calculated by substituting `t = t_m`

in equation `y = (v_o sin theta_o ) t – 1/2 g t^2`.

`y=h_m=(v_o sin theta_o)((v_o sin theta_o)/g)-g/2 ((v_o sin theta_o)/g)^2`

`color(red)(h_m=(v_o sin theta_o)^2 /(2g))`

The horizontal distance travelled by a projectile from its initial position `(x = y = 0)` to the position where it passes `y = 0` during its fall is called the horizontal range, R.

It is the distance travelled during the time of flight `T_f`.

Therefore, the range `R` is
`color(red)(R=(v_o cos theta_o) (t_f))`

`color(blue)(R=(v_o cos theta_o) ( 2 v_o sin theta_o) / g)`

`color(red)(R=(v_o^2 sin 2theta_o)/g)`

• `R` is maximum when `sin 2theta_o` is maximum, i.e., when `theta_o = 45^o`

`color(green)(ul"The maximum horizontal range is"` `R_m=(v_o^2)/g)`