Up-till 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 vx. 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 vx, 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 velocity 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?
Application To Ballistic Missiles
A ballistic flight is that in which a projectile is given an initial push and is then allowed to move freely due to inertia and under the action of gravity. An un-powered and un-guided missile is called a ballistic missile and the path followed by it is called ballistic trajectory.
As discussed before, a ballistic missile moves in a way that is the result of the superposition of two independent motion: a straight line inertial flight in the direction of the launch and a vertical gravity fall. By law of inertia, an object should sail straight off in the direction thrown, at constant speed equal to its initial speed particularly in empty space. But the downward force of gravity will alter straight path into a curved trajectory. For short ranges and flat Earth approximation, the trajectory is parabolic but the dragless ballistic trajectory for spherical Earth should actually be elliptical. At high speed and for long trajectories the air friction is not negligible and some times the force of air friction is more than gravity. It affects both horizontal as well as vertical motions. Therefore, it is completely unrealistic to neglect the aerodynamic forces.
The shooting of a missile on a selected distant spot is a major element of warfare. It undergoes complicated motions due to air friction and wind etc. consequently the angle of projection can not be found by the geometry of the situation at the moment of launching. The actual flights of missiles are worked out to high degrees of precision and the result were contained in tabular form. The modified equation of trajectory is too complicated to be discussed here. The ballistic missiles are useful only for short ranges. For long ranges and greater precision, powered and remote control guided missiles are used.
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Do You Know? |
 For an angle less than 45⁰, the height reached by the projectile and the range both will be less. When the angle of projectile is larger than 45⁰, the height attained will be more but the range is again less. |
Example 3.7: A ball is thrown with a speed of 30 ms-1 in a direction 30⁰ above the horizon. Determine the height to which it rises, the time of flight and the horizontal range.
Solution: initially
Example 3.8: In example 3.7 calculate the maximum range and the height reached by the ball if the angles of projection are (i) 45⁰ (ii) 60⁰.
Solution:
(i) Using the equation for height and range we have
(ii) Using the equation for height and range we have
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For your information |
 In the presence of air friction the trajectory of a high speed projectile fall short of a parabolic path. |