Projectile Motion
The path that an object takes when thrown at an angle other than 90 degrees from a horizontal point is known as a trajectory, the object is known as a projectile, and the motion is defined as the projectile motion. Football, baseball, cricket ball, and any other object are instances of projectile motion.
There are many types of motion around us. Projectile motion is a form of motion experienced by an object or particle that is thrown near the Earth's surface and moves along a curved path under the action of gravity only (in particular, the effects of air resistance are assumed to be negligible). Such an object is called a projectile and the curved path with which the projectile travels is what is termed as trajectory.
Formulae related to a projectile that is launched at an oblique angle with respect to the horizontal and whose motion is referred with respect to the horizontal are:
(v θ is the velocity of projection,
Formulas and Concepts of Projectile Motion
The projectile motion is divided into two parts: a horizontal motion with no acceleration and a vertical motion with constant acceleration due to gravity.
Consider the following example of a ball being launched at an angle from point O to the horizontal x-axis with an initial velocity of u:
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Point O is defined as the point of projection.
θ is defined as the angle of projection and
OB = Horizontal Range.
H is defined as the height of the particle.
The time of flight refers to the amount of time it takes a particle to travel from point A to point B.
Differential equations of motion can be used to discover various projectile motion parameters.
The linear equation of motion is:
v = u + at
S = ut + 1/2(at2)
v2 = u2 + 2aS
Apply the above equation for projectile motion, the equation will now be,
v = u – gt
S = ut – 1/2(gt2)
v2 = u2 – 2gS
Here,
u = initial velocity
v = Final velocity
g = Acceleration due to gravity (Taking it negative because gravity always work downward)
S = Displacement
t = Time
Time of Flight
It is the total amount of time the projectile remains in the air.
In Y direction total displacement (Sy) = 0.
Taking motion in Y direction, Sy = uyt – 1/2(gt2)
(Here, uy = u sinθ and Sy = 0)
i.e. 0 = usinθ – 1/2(gt2)
t = 2usinθ/g
Time of Flight(t) = 2usinθ/g
Maximum Height
It is the particle's highest point (point A). The vertical component of the velocity (Vy) will be zero when the ball reaches point A.
That is, 0 = (usinθ)2 – 2gHmax
( S = Hmax, vy = 0 and uy = u sin θ )
The Maximum Height of the projectile is:
Maximum Height (Hmax) = u2sin2θ/2g
Horizontal Range
It is defined as the horizontal distance covered to the maximum distance possible.
The horizontal range is a distance (OB) is:
OB = Horizontal component of velocity(ux) * Total time(t)
(ux = u cosθ and t = 2usinθ/g)
That is, Range(R) = ucosθ * 2usinθ/g
Horizontal Range of the projectile is:
Horizontal Range(R) = u2sin2θ/g
( sin2θ = 2cosθsinθ )
The Equation of Trajectory
The trajectory equation is the path taken by a particle during projectile motion. The following is the equation:
y = x tanθ – gx2/2u2cos2θ
Because this equation is similar to the parabola (y = ax + bx2), it is concluded that that projectile motion is always parabolic in character.
Solved Example
Question: A guy can swim at a speed of 4.0 km/h in still water. How long does it take him to traverse a 1.0 km wide river at 3.0 km/h while keeping his strokes parallel to the current? After he reaches the opposite bank, how far down the river does he go?
Sol:
Speed of the guy = 4 km/h and width of the river = 1 km
Therefore, the time is taken to cross the river = Width of the river / Speed of the river
= 1/4 h
= 1/4 × 60 = 15 minutes
The Speed of the river, vr is equal to 3 km/h
Thus, distance covered with flow of the river = vr × t
= 3 × 1/3 = 1/4
= 3/4 × 1000 = 750 m
Question: If the horizontal range of a projectile is 4 times the maximum height attained by it, then the angle of projection is:
Options:
(a) 450
(b) 300
(c) 600
(d) 150
Answer: (a)
FAQs on Projectile Motion Formula
1. What are some examples of projectile motion applications?
Projectile Motion can be used in a variety of ways. In modern life, a rocket or missile is a more complicated sort of projectile application. Athletes frequently use projectiles in events such as the javelin throw, shot put, discus, and hammer throw, among others. It is used in archery and shooting.
2. What are some examples of two-dimensional motion?
Two-dimensional motion can be seen in the following examples:
Throwing a ball
A billiard ball's movement on a billiard table
An artillery shell's firing motion
The earth's rotation around the sun
A projectile motion is created when a ball is hurled from a moving vehicle
A fielder in a cricket match sends the ball towards the wickets
A bullet strikes a long-range target
3. What are the physical quantities that will remain constant for a projectile?
At any time, vx=v cos θ, ax = 0
vy= v sin θ−gt, ay=−g
Since vy is changing, so (with vx constant)
Kinetic Energy will change with vy.
As Potential Energy is a function of height, it will change as well.
Therefore, only acceleration and the horizontal components of velocity remain constant.
4. Do projectile motion get affected by gravity?
Vertical acceleration is caused by gravity's influence on the projectile's vertical velocity. The tendency for any moving object to maintain a constant velocity causes the projectile to move horizontally. Due to the lack of horizontal forces, a bullet maintains a constant horizontal velocity. No horizontal forces are required to keep a bullet moving horizontally. The only force operating on a projectile is gravity!
5. How can CBSE Class 10 Physics students receive free Projectile Motion questions and answers on Vedantu?
Vedantu has carefully created Projectile Motion questions and answers solutions for class 10 that can assist students to comprehend the concepts and learn how to answer correctly in board exams. Free PDF Solutions are also available for students to download.
