Projectile Motion Lecture VideoParabolic Motion of ProjectilesBasketball PhysicsSolved Example
What is Projectile?
Projectile is any object thrown into space upon which the only acting force is the gravity. In other words, the primary force acting on a projectile is gravity. This doesn’t necessarily mean that the other forces do not act on it, just that their effect is minimal compared to gravity. The path followed by a projectile is known as a trajectory. A baseball batted or thrown and the instant the bullet exits the barrel of a gun are all examples of projectile.
What is Projectile Motion?
When a particle is thrown obliquely near the earthâ€™s surface, it moves along a curved path under constant acceleration that is directed towards the centre of the earth (we assume that the particle remains close to the surface of the earth). The path of such a particle is called a projectile and the motion is calledÂ projectile motion. Air resistance to the motion of the body is to be assumed absent in projectile motion.
In a Projectile Motion, there are two simultaneous independent rectilinear motions:
 Along xaxis: uniform velocity, responsible for the horizontal (forward) motion of the particle.
 Along yaxis: uniform acceleration, responsible for the vertical (downwards) motion of the particle.
Accelerations in the horizontal projectile motion and vertical projectile motion of a particle: When a particle is projected in the air with some speed, the only force acting on it during its time in the air is the acceleration due to gravity (g). This acceleration acts vertically downward. There is no acceleration in the horizontal direction which means that the velocity of the particle in the horizontal direction remains constant.
If you prefer a video explanation of projectile motion, here is an engaging video for you.
Parabolic Motion of Projectiles
Let us consider a ball projected at an angle Î¸ with respect to horizontal xaxis with the initial velocity u as shown below:
The point O is called the point of projection; Î¸ is the angle of projection and OB = Horizontal Range or Simply Range. The total time taken by the particle from reaching O to B is called the time of flight.
For finding different parameters related to projectile motion, we can make use of different equations of motions:
Total Time of Flight: Resultant displacement (s) = 0 in Vertical direction. Therefore, by using the Equation of motion:
gt2 = 2(uyt â€“ sy) [Here, uy = u sin Î¸ and sy = 0]
i.e. gt2 = 2t Ã— u sin Î¸
Therefore, the total time of flight (t):
\(Total\,Time\,of\,Flight\,(t)=\frac{2u\sin\Theta}{g}\)
Horizontal Range: Horizontal Range (OA) = Horizontal component of velocity (ux) Ã— Total Flight Time (t)
R = u cos Î¸ Ã— 2uÃ—sinÎ¸g
Therefore in a projectile motion the Horizontal Range is given by (R):
\(Horizontal\,Range\,(R)=\frac{u^2\sin 2\Theta }{g}\)
Maximum Height: It is the highest point of the trajectory (point A). When the ball is at point A, the vertical component of the velocity will be zero. i.e. 0 = (u sin Î¸)2 â€“ 2g Hmax [s = Hmax , v = 0 and u = u sin Î¸]
Therefore in a projectile motion the Maximum Height is given by (Hmax):
\(Maximum\,Height\,(H_{max})=\frac{u^2\sin^2\Theta }{2g}\)
The equation of Trajectory: Let, the position of the ball at any instant (t) be M (x, y). Now, from Equations of Motion:
x = t Ã— u cos Î¸ . . . . . . (1)
y = u sin Î¸ Ã— t â€“ 12Ã—t2g. . . . . . (2)
On substituting Equation (1) in Equation (2):
\(Equation\,of\,Trajectory\,(y)=\frac{x\tan \Theta gx^2}{2u^2\cos^2\Theta }\)
This is the Equation of Trajectory in a projectile motion, and it proves that the projectile motion is always parabolic in nature.
Basketball Physics
We know that projectile motion is a type of twodimensional motion or motion in a plane. It is assumed that the only force acting on a projectile (the object experiencing projectile motion) is the force due to gravity. But how can we define projectile motion in the real world? How are the concepts of projectile motion applicable to daily life? Let us see some reallife examples of projectile motion in two dimensions.
All of us know about basketball. To score a basket, the player jumps a little and throws the ball in the basket. The motion of the ball is in the form of a projectile. Hence it is referred to as projectile motion. What advantage does jumping give to their chances of scoring a basket? Now apart from basketballs, if we throw a cricket ball, a stone in a river, a javelin throw, an angry bird, a football or a bullet, all these motions have one thing in common. They all show a projectile motion. And that is, the moment they are released, there is only one force acting on them the gravity. It pulls them downwards, thus giving all of them an equal impartial acceleration of.
It implies that if something is being thrown in the air, it can easily be predicted how long the projectile will be in the air and at what distance from the initial point it will hit the ground. If the air resistance is neglected, there would be no acceleration in the horizontal direction. This implies that as long as a body is thrown near the surface, the motion of the body can be considered as a twodimensional motion, with acceleration only in one direction. But how can it be concluded that a body thrown in air follows a twodimensional path? To understand this, let us assume a ball that is rolling as shown below:
Thus, it can be concluded that the minimum number of coordinates required to completely define the motion of a body determines the dimension of its motion.
Solved Example For You
An object is launched at a velocity of 40 m/s in a direction making an angle of 50Â° upward with the horizontal.

 What is the maximum height reached by the object?
 What is the total flight time (between launch and touching the ground) of the object?
 What is the horizontal range (maximum x above ground) of the object?
Solution:

 The velocity components \(V_{x}\) and \(V_{y}\) are given by the formula:
In the given problem, \(V_{0} = 40m/s\), ? = 50Â° and g is \(9.8\,m^{2}\) The height of the projectile is given by the component y, and it reaches its maximum value when the component \(V_{y}\) is equal to zero. That is when the projectile changes from moving upward to moving downward.
Substituting and solving for t, we get\(t=\frac{V_{0}\sin(?)}{g} = \frac{40\sin (50^{\circ})}{9.8}=3.12\,seconds\)To find the maximum height, substitute t in the equation y, we get
\(y=\frac{V_0\sin ?}{3.12}\frac{1}{2}(9.8)(3.12^2)\)Solving, we get
\(y=\frac{40\sin50^{\circ}}{3.12}\frac{1}{2}(9.8)(3.12^2)=47.9\,meters\)The maximum height reached by the object is 47.9 meters
 The time of flight is the interval between when the projectile is launched (t_{1}) and when the projectile touches the ground (t_{2}).
Hence,
\(V_0\sin(?)t\frac{1}{2}gt^2\) Solve for t
\(t(V_0\sin(?)(\frac{1}{2}gt)=0\) Solving, we get two soluions as follows:
\(t=t_1=0\) and \(t=t_2=2V_0sin(?)/g\)Time of flight can be calculated as follows: \(Time\,of\,Flight=2(20)\sin (?)/g=6.25\,seconds\)
 The velocity components \(V_{x}\) and \(V_{y}\) are given by the formula:
 Horizontal Range is the horizontal distance given by x at t = t_{2}.
\(Range=x(t_2)=(V_0\cos(?)V_0\sin (?))/g=(V_0)^2\sin(2?)/g=40^2\sin(2(50^{\circ}))/9.8=6.25\,meters\)
Stay tuned with BYJU’S to learn more about projectile motion and its applications.
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