2.6 Projectile Motion

2.6 Projectile Motion

Uptill now we have been studying the motion of a particle along a straight line i.e., motion in one dimension. Now we consider the motion of a ball, when it is thrown horizontally from certain height. It is observed that the ball travels forward as well as falls downward, until it strikes something such as ground. Suppose that the ball leaves the hand of the thrower at point $A$ (Fig 2.8-a) and that its velocity at that instant is completely horizontal. Let this velocity be $v_{x}$. According to Newton's first law of motion, there will be no acceleration in horizontal direction, unless a horizontally directed force acts on the ball. Ignoring the air friction, only force acting on the ball during flight is the force of gravity. There is no horizontal force acting on it. So, its horizontal velocity will remain unchanged and will be $v_{x}$, until the ball hits the ground. The horizontal motion of ball is simple. The ball moves with constant horizontal velocity component. Hence, horizontal distance $x$ is given by

The vertical motion of the ball is also not complicated. It will accelerate downward under the force of gravity and hence $a = g$. This vertical motion is the same as for a freely falling body. Since initial vertical velocity is zero, hence, vertical distance $y$, using Eq. 2.14 is given by

It is not necessary that an object should be thrown with some initial velocity in the horizontal direction. A football kicked off by a player; a ball thrown by a cricketer and a missile fired from a launching pad, all projected at some angles with the horizontal, are called projectiles.

Projectile motion is two dimensional motion under constant acceleration due to gravity.

In such cases, the motion of a projectile can be studied easily by resolving it into horizontal and vertical components which are independent of each other. Suppose that a projectile is fired in a direction angle $\theta$ with the horizontal by velocity $v_{i}$, as shown in Fig.2.8(b). Let components of velocity $v_{x}$ along the horizontal and vertical directions be $v_{i}\cos\theta$ and $v_{i}\sin\theta$, respectively. The horizontal acceleration is $a_{x} = 0$ because we have neglected air resistance and no other force is acting along this direction, whereas the vertical acceleration is $a_{y} = g$. Hence, the horizontal component $v_{x}$ remains constant and at any time $t$, we have

Now we consider the vertical motion. The initial vertical component of the velocity is $v_{i}\sin\theta$ in the upward direction.

The vertical component $v_{y}$ at any instant $t$ can be determined by considering the upward motion of projectile as free fall motion ($a_{y} = -g$). Using 1st equation of motion:

The magnitude of velocity at any instant is:

The angle $\phi$ which this resultant velocity makes with the horizontal can be found from

In projectile motion one may wish to determine the height to which the projectile rises, the time of flight and horizontal range. These are described below.