Sales Toll Free No: 1-855-666-7446

Projectile Motion


You may have a habit of playing a base ball. you could see that when you kick it the ball takes the curved path and later returns to the ground.
Projectile Motion
This curved path what any object takes when thrown in the air is what we call as Projectile motion!

Projectile Motion Definition

Back to Top
Its a motion that a object undertakes when projected into the air at an angle that leads to the curved path and reaches the ground.
Projectile Motions
This path taken by the projectile is called as trajectory of projectile. It is acted upon by the acceleration due to the gravity that gives it a curved path in a vertical plane.

Projectile Motion Equations

Back to Top
Observing the above projectile motion figure we get to know that
Horizontal distance x = Vx t
Horizontal velocity Vx = Vxo
vertical distance y = Vyo t - $\frac{1}{2}$ gt2
Vertical velocity Vy = Vyo - gt

Here Vx is the velocity along x-axis,
Vxo is the initial velocity along x-axis,
Vy is the velocity along y-axis,
Vyo is the initial velocity along y-axis,
g is the acceleration due to gravity,
t is the time taken.
The equation of projectile is given by
y = $\frac{1}{2}$ (at2) + Vo t + yo
Here y = Height of the projectile,
t = time
a = acceleration of the projectile due to gravity
Vo = Initial velocity of projectile
yo = Initial height of projectile.

Range Equation for Projectile Motion

Back to Top
The Horizontal range is the horizontal distance traveled by the projectile. It is given as R in figure. It is the distance traveled by the projectile in a time equal to time of flight tf.
Velocity v = u cos $\theta$
$\therefore$ R = (u cos $\theta$) tf
The time of flight is tf = $\frac{2u sin \theta}{g}$

R = (u cos $\theta$) $\frac{2 u sin \theta}{g}$

= $\frac{2u^2 cos \theta sin \theta}{g}$

= $\frac{u^2 (2 sin \theta cos \theta)}{g}$

= $\frac{u^2 sin 2 \theta}{g}$

R = $\frac{u^2 sin 2 \theta}{g}$

Projectile Motion Maximum Height Equation

Back to Top
The Maximum height of the projectile motion is the elevation to the highest point reached by the projectile during its flight. It is given as
At the highest vertical point, Initial velocity = u sin $\theta$
Final velocity v = 0, displacement = H and acceleration a = -g
V2 = u2 + 2as
0 = (u sin $\theta$)2 - 2gH
2gH = u2 sin2 $\theta$

Hence the Height is

H = $\frac{u^2 sin^2 \theta}{2g}$.

Horizontal Projectile Motion

Back to Top
A body thrown horizontally from a certain height H moves with a velocity u covers a horizontal distance x in time t.
Horizontal Projectile
In vertical motion the height H is given by
H = $\frac{1}{2}$ gt2

Projectile Motion with Air resistance

Back to Top
The only force acting in projectile motion is gravitational force. Hence the drag by the air resistance is considered. The projectile motion by considering the air resistance is a very complex thing. It can be solved only by using numerical integration. The force due to air resistance is opposite in direction to the vector velocity in a particular time.

Projectile Motion Problems

Back to Top
Here are some sample problems on projectile motion you can go through it :

Solved Examples

Question 1: A bullet is fired from a gun at a rate of 1500 m/s that makes the angle with the horizontal. Calculate
(i) Time taken to reach the highest point
(ii) Total time of flight.
If u is the initial velocity of the bullet and $\theta$ is the angle of projection
Time taken to reach the highest point is t = $\frac{u sin \theta}{g}$
                                                            = $\frac{1500 \times sin 30^o}{9.8}$
                                                            = 76.53 s

Time of flight T = 2t = 2 $\times$ 76.53 s = 153.06 s.

Question 2: A horizontal range of a projectile is 4 $\sqrt{3}$ times its maximum height. Find its angle of projection.
Given: Range of projectile R = 4 $\sqrt{3}$ $\frac{2u^2 sin \theta\ cos \theta}{g}$
                                         = 4 $\sqrt{3}$ $\times$ $\frac{u^2 sin^2 \theta}{2g}$

=> $\frac{sin \theta}{cos \theta}$ = tan $\theta$ = $\frac{1}{\sqrt{3}}$
$\therefore$ $\theta$ = 30o.