# Projectile Motion Part 1 (F.Sc-Physics-Chapter 3.12)

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 downwards, until it strikes something. Suppose that the ball leaves the hand of the thrower at point a (Fig. 3.16 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 something. 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. 3.7, 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 θ with the horizontal by velocity Vi as shown in Fig. 3.16 (b). let components of velocity Vi along the horizontal and vertical directions Vi cos θ and vi sin θ respectively. The horizontal acceleration is ax = 0 because we have neglected air resistance and no other force is acting along this direction whereas vertical acceleration a y = g. hence , the horizontal component Vix remains constant and at any time t, we have

Now we consider the vertical motion. The initial vertical component of the velocity is sin θ in the upward direction. Using Eq. 3.5 the vertical component of the velocity at any instant t is given by

The magnitude of velocity at any instant is

The angle ф 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.

**Intrusting Information **

A photograph of two balls released simultaneously from a mechanism that allows one ball to drop freely while the other is projected horizontally, at any time the two balls are at the same level, i.e., their vertical displacements are equal.

**Height of the Projectile**

In order to determine the maximum height the projectile

**Time of flight**

The time taken by the body to cove the distance from the place of its projection to the place where it hits the ground at the same level is called the time of flight.

This can be obtained by taking S = h = 0, because the body goes up and comes back to same level, thus covering no vertical distance. If the body is projecting with velocity v making angle θ with a horizontal, then its vertical component will be Vi sin θ. Hence the equation is

Where t is the time of flight of the projectile when it is projected from the ground as shown in Fig. 3.16 (b).

**Range of the projectile **

Maximum distance which a projectile covers in the horizontal direction is called the range of the projectile.

To determine the range R of the projectile, we multiply the horizontal component of the velicity of projection with total time taken by the body after leaving the point of projection. Thus

But, 2 sin θ cos θ = sin2 θ, thus the range of the projectile depends upon the velocity of projection and the angle of projection.

For the range R to be maximum, the factor sin2θ should have maximum value which is 1 when 2 θ = 90⁰ or θ = 45⁰

**Point to ponder**

Water is projected from two rubber pipes at the same speed-from one at an angle of 30⁰ and from the other at 60⁰. Why are the ranges equal?