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Equations of motion in horizontal direction

The force due to gravity has no component in horizontal direction. Since gravity is the only force acting on the projectile, this means that the motion in horizontal direction is not accelerated. Therefore, the motion in horizontal direction is an uniform motion. This implies that the component of velocity in x-direction is constant. As such, the position or displacement in x-direction at a given time “t” is :

x = u x t

Projectile motion

Horizontal displacement at a given time

This equation gives the value of horizontal position or displacement at any given instant.

Displacement of projectile

The displacement of projectile is obtained by vector addition of displacements in x and y direction. The magnitude of displacement of the projectile from the origin at any given instant is :

Displacement, OP = ( x 2 + y 2 )

Displacement in projectile motion

The angle that displacement vector subtends on x-axis is :

tan α = y x

Velocity of projectile

The velocity of projectile is obtained by vector addition of velocities in x and y direction. Since component velocities are mutually perpendicular to each other, we can find magnitude of velocity of the projectile at any given instant, applying Pythagoras theorem :

v = ( v x 2 + v y 2 )

Velocity of a projectile

The angle that the resultant velocity subtends on x-axis is :

tan β = v y v x

Problem : A ball is projected upwards with a velocity of 60 m/s at an angle 60° to the vertical. Find the velocity of the projectile after 1 second.

Solution : In order to find velocity of the projectile, we need to know the velocity in vertical and horizontal direction. Now, initial velocities in the two directions are (Note that the angle of projection is given in relation to vertical direction.):

u x = u sin θ = 60 sin 60 ° = 60 x 3 2 = 30 3 m / s u y = u cos θ = 60 cos 60 ° = 60 x 1 2 = 30 m / s

Now, velocity in horizontal direction is constant as there is no component of acceleration in this direction. Hence, velocity after "1" second is :

v x = u x = 30 3 m / s

On the other hand, the velocity in vertical direction is obtained, using equation of motion as :

v y = u y - g t v y = 30 - 10 x 1 v y = 20 m / s

The resultant velocity, v, is given by :

v = ( v x 2 + v y 2 ) v = { ( 30 3 ) 2 + ( 20 ) 2 } = ( 900 x 3 + 400 ) = 55.68 m / s

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Equation of the path of projectile

Equation of projectile path is a relationship between “x” and “y”. The x and y – coordinates are given by equations,

y = u y t - 1 2 g t 2 x = u x t

Eliminating “t” from two equations, we have :

y = u y x u x - g x 2 2 u x 2

For a given initial velocity and angle of projection, the equation reduces to the form of y = A x + B x 2 , where A and B are constants. The equation of “y” in “x” is the equation of parabola. Hence, path of the projectile motion is a parabola. Also, putting expressions for initial velocity components u x = u cos θ and u y = u sin θ , we have :

y = ( u sin θ ) x u cos θ - g x 2 2 u 2 cos 2 θ y = x tan θ - g x 2 2 u 2 cos 2 θ

Some other forms of the equation of projectile are :

y = x tan θ - g x 2 sec 2 θ 2 u 2

y = x tan θ - g x 2 ( 1 + tan 2 θ ) 2 u 2


A projectile with initial velocity 2 i + j is thrown in air (neglect air resistance). The velocity of the projectile before striking the ground is (consider g = 10 m / s 2 ) :

(a) i + 2 j (b) 2 i j (c) i – 2 j (d) 2 i – 2 j

The vertical component of velocity of the projectile on return to the ground is equal in magnitude to the vertical component of velocity of projection, but opposite in direction. On the other hand, horizontal component of velocity remains unaltered. Hence, we can obtain velocity on the return to the ground by simply changing the sign of vertical component in the component expression of velocity of projection.

Projectile motion

Components of velocities

v = 2 i - j

Hence, option (b) is correct.

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Questions & Answers

how did you get the value of 2000N.What calculations are needed to arrive at it
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can anyone tell who founded equations of motion !?
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n=a+b/T² find the linear express
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Sultan Reply
Moment of inertia of a bar in terms of perpendicular axis theorem
Sultan Reply
How should i know when to add/subtract the velocities and when to use the Pythagoras theorem?
Yara Reply
Centre of mass of two uniform rods of same length but made of different materials and kept at L-shape meeting point is origin of coordinate
Rama Reply
A balloon is released from the ground which rises vertically up with acceleration 1.4m/sec^2.a ball is released from the balloon 20 second after the balloon has left the ground. The maximum height reached by the ball from the ground is
Lucky Reply
work done by frictional force formula
Sudeer Reply
Misthu Reply
Why are we takingspherical surface area in case of solid sphere
Saswat Reply
In all situatuons, what can I generalize?
Cart Reply
the body travels the distance of d=( 14+- 0.2)m in t=( 4.0 +- 0.3) s calculate it's velocity with error limit find Percentage error
Clinton Reply
Explain it ?Fy=?sN?mg=0?N=mg?s
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