A projectile is something which remains in flight after being thrown and after sometimes it comes back to the ground. a stone thrown by someone, a bullet fired from a gun are the examples of projectile.

_{0}of the projectile can be resolved into two components along the two perpendicular axes, viz,

**v**along the x-axis and

_{o}cosθ_{o}**v**along the y axis.

_{ o}sinθ_{ o}Here it is assumed that the air resistance is negligible on the motion of the projectile. Suppose that the projectile is launched with velocity vo that makes an angle θo with the x-axis.

Here we have to assume that the only force acting on the projectile is the gravitational force, due to which the acceleration of it is g which is acting along the y-axis in the downward direction.

So acceleration is

As the figure shows, the point of projection, i.e, the point from which it is projected is the origin O. After any time t the distance traveled along the axes is

** x = v _{0x}t = v_{0}cosθ_{0}t**…………………………….(1)

And

………(2)

Now the path followed by the projectile can be easily calculated by eliminating t from both the equation (1) and (2). Hence we get from eq. (1)

From eq. (2)

This is the equation of path that has been followed by the projectile. This equation is very similar to the standard equation

** **which is an equation of parabola, where a and b are constants. Therefore, the path will be a parabolic one.

Now as the projectile reaches its maximum height the y component of velocity becomes zero and it starts moving downwards. The time taken by it to reach this maximum height is calculated by putting velocity vy = 0

Now,

Here tm is the required time. Hence

Maximum height attended by it can be easily calculated by using tm in equation (2)

After putting the value we get

And the time of flight is given by putting y=0

Here we can have either t=0 or

Since t cannot be zero, time of flight is

The path travelled by it along the x-axis or travelled horizontally, I.e, the distance travelled from initial position O(0,0) to the final position where y becomes zero again is known as range or horizontal range. The horizontal component of velocity is only responsible for this range, hence the distance travelled by the projectile due to v0x in time Tf is the range. Thus,

Range

Therefore

Here the point to remember is that at , the value if R will be maximum.

Here one interesting statement made by Galileo that for elevations which exceed or fell short of 45° by an equal amount, the ranges of both of them are equal.

It can be understood easily. For an angle , we have and , the value of is and respectively. Hence and will have the same value of Hence the ranges will be same.

**PROBLEM SOLVING VIDEO ON PROJECTILE & ITS MOTION**