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.
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θ is defined as the angle of projection and
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OB = Horizontal Range.
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H is defined as the height of the particle.
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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)