Kinematics of Projectile Motion

Until now we read about motion along a straight line. But that is not always so in real life.
 
                                           

In the picture above, for example, the motor cyclist is travelling in 2 dimensions because he is moving in a plane, not along a straight line.

Similarly, when you throw a ball at an angle to the horizontal, the motion of the ball is in 2 dimensions.

                               
 
Motion in 2 dimensions under constant acceleration is generally termed projectile motion.

Understanding Projectile Motion

Let us take the common case of a ball being projected at an angle θ with the horizontal under the influence of gravity. This is an example of projectile motion.
 
                                  

Let the initial velocity of the ball be u.

Now we want to find the velocity of this ball at any instant after the instant of projection (t = 0). We want to find the maximum height reached by the ball. We want to find the net horizontal distance covered by the ball when it touches the ground again. We want to find the net time taken by the ball in reaching the ground again.

How do we do all this?

In case of straight line motion with constant acceleration, we applied the following equations:

v = u + at

v² = u² + 2as even

s = ut + ½ at²

We cannot apply these equations directly in case of projectile motion because the ball keeps changing its direction. We generally apply the equations above in one single direction.

So we do something interesting.

We resolve the velocity of the ball into 2 components – horizontal and vertical. And we interpret projectile motion as two simultaneous motions of the ball – one along the horizontal and one along the vertical.

Then we can apply the 3 equations of motion along the x and y axis for the motion of the ball.

Along the x axis

 

Now, along the x axis, the velocity of the ball is ucosθ. There is no acceleration along the x axis, so the ball moves with constant velocity along this direction. The next displacement of the ball along the x axis will therefore be ucosθt.
 
                           

Along the y axis



Along the y axis, the velocity is not constant. The initial velocity is usinθ, but gravity is acting downwards with an acceleration g.

Hence, the velocity along the y axis at any instant is usinθ – gt.
Similarly the net displacement along the y axis is usinθt – ½gt.²

                   

Therefore, this is how we can represent the net motion of the ball along both the x and y axis.

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