# What is projectile motion?

An object that is in flight after being thrown or projected is said to be in projectile motion. In real life, we can see one such example as the cricket ball in air.

The motion of a projectile may be thought of as the result of two independent, simultaneously occurring motions. One component is along the horizontal direction without any acceleration and the other along the vertical direction with constant acceleration due to the force of gravity. In this article, we shall assume that the air resistance has a negligible effect on the motion of the projectile.

## Some equations for projectile motion:

Suppose that the projectile is launched with velocity v that makes an angle θ with the x-axis. After the object has been projected, the acceleration acting on it is due to gravity which is directed vertically downwards:

a(combined) = −g Eqn (1)
ax= 0, ay = –g Eqn (2)

The components of initial velocity v are :
vx = vcosθ Eqn (3)
vy = vsinθ Eqn (4)

If we take the initial position to be the origin of the reference frame, we have x & y= 0;

Then at any time t, displacement can be obtained using:
x = (vx)t = (vcosθ)t
y = (vsinθ)t – (½)gt2 Eqn (5)

The components of velocity at time t can be obtained using:
vx = vcosθ
vy = vsinθ – gt Eqn (6)

Using these & establishing relation between y and x we obtain:
y=(tanθ)x – gx2/(4v2cos2θ) Eqn (7)

Notice that the choice of mutually perpendicular x and y directions for the analysis of the projectile motion has resulted in a simplification. One of the components of velocity, i.e. x-component remains constant throughout the motion and, only the y-component changes, like an object in free fall in the vertical direction. Note that at the point of maximum height, θ=0 and therefore, vy = 0.

### Time Of Flight

The total time Tf during which the projectile is in flight can be obtained by substituting y = 0, i.e. Zero displacement in vertical direction, in Eqn (7).

We obtain, Tf = 2vsinθ/g Eqn (8)

Tf is known as the time of flight of the projectile. We note that Tf = 2tm ,where tm is the time take to reach maximum height. It is intuitive because of the symmetry of the parabolic path.

### Maximum Height

The maximum height Hm reached by the projectile can be calculated by substituting t = tm in Eqn (7):

Hm=((v2sinθ))/(2g) Eqn (9)

### Range Of the Projectile

The horizontal distance covered by a projectile from its initial position (x = y = 0) to the position where it passes y = 0 during its fall is called the horizontal range, R. It is the distance travelled during the time of flight Tf. Therefore, the range R is:

R = (vcosθ).(Tf) =(vcosθ)(2vsinθ)/g , which simplifies to:
R =(v2sin(2θ))/g Eqn (10)

### Suggestions:

These are some general points to keep in mind.

• Projectile motions can be solved easily when solved as two separate motions.
• Use the fundamental equations whenever tackling position based problems.
• Clear your concepts of vectors before attempting to understand this topic.
• Moreover, for building concepts & advanced practice, solve Concepts of Physics by H.C.Verma.

Projectile motion is just the opening bit, before the vast cavern called Mechanics. Read how to attempt Mechanics for JEE Advanced here.

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