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.

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.
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.
Observing the above projectile motion figure we get to know thatHorizontal distance x = V_{x} t
Horizontal velocity V_{x} = V_{xo}
vertical distance y = V_{yo} t  $\frac{1}{2}$ gt^{2}
Vertical velocity V_{y} = V_{yo}  gt
Here V_{x} is the velocity along xaxis,Horizontal velocity V_{x} = V_{xo}
vertical distance y = V_{yo} t  $\frac{1}{2}$ gt^{2}
Vertical velocity V_{y} = V_{yo}  gt
V_{xo} is the initial velocity along xaxis,
V_{y} is the velocity along yaxis,
V_{yo} is the initial velocity along yaxis,
g is the acceleration due to gravity,
t is the time taken.
The equation of projectile is given by
y = $\frac{1}{2}$ (at^{2}) + V_{o} t + y_{o}
_{}Here y = Height of the projectile,t = time
a = acceleration of the projectile due to gravity
V_{o} = Initial velocity of projectile
y_{o} = Initial height of projectile.
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 t_{f}.
Velocity v = u cos $\theta$
$\therefore$ R = (u cos $\theta$) t_{f}
The time of flight is t_{f }= $\frac{2u sin \theta}{g}$
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 = $\frac{u^2 sin 2 \theta}{g}$
$\therefore$ R = (u cos $\theta$) t_{f}
The time of flight is t_{f }= $\frac{2u sin \theta}{g}$
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 = $\frac{u^2 sin 2 \theta}{g}$
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
V^{2} = u^{2} + 2as
0 = (u sin $\theta$)^{2}  2gH
2gH = u^{2} sin^{2} $\theta$
Hence the Height is
H = $\frac{u^2 sin^2 \theta}{2g}$.
Final velocity v = 0, displacement = H and acceleration a = g
V^{2} = u^{2} + 2as
0 = (u sin $\theta$)^{2}  2gH
2gH = u^{2} sin^{2} $\theta$
Hence the Height is
H = $\frac{u^2 sin^2 \theta}{2g}$.
A body thrown horizontally from a certain height H moves with a velocity u covers a horizontal distance x in time t.
In vertical motion the height H is given by
In vertical motion the height H is given by
H = $\frac{1}{2}$ gt^{2}^{}
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.
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.
Solution:
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.
(i) Time taken to reach the highest point
(ii) Total time of flight.
Solution:
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.
Solution:
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 $\theta$ = $\frac{1}{\sqrt{3}}$
Solution:
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 $\theta$ = $\frac{1}{\sqrt{3}}$
$\therefore$ $\theta$ = 30^{o}.