Table of contents 
What is Projectile? 
Frequently Asked Questions (FAQs) 
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When a particle is thrown obliquely near the earth’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 called projectile motion.
In a Projectile Motion, there are two simultaneous independent rectilinear motions:
Let us consider a ball projected at an angle θ with respect to the horizontal xaxis with the initial velocity u as shown below:
For finding different parameters related to projectile motion, we can make use of differential equations of motions:
Equation of Trajectory If x and y are the coordinates of particle after time t:
From the equation (i) by putting the value of t, as a function of x, in equation (ii), we get:
The above equation is called the equation of trajectory. As the equation represents a parabola. Thus, the trajectory (or the path) of a projectile is a parabola.
Tip: Here, u, θ, x and y are four variables. If any three quantities, as mentioned, are known then the fourth quantity can be solved directly
The time taken by a projectile to return its initial elevation after projection is known as time of flight.
It is denoted by (T) and given by:
Horizontal Range (OA) = Horizontal component of velocity (u_{x}) X Total Flight Time (t)
R = u cos θ × (2usinθ)/g
Where,
Therefore, in a projectile motion the Horizontal Range is given by (R):
Q.1. An object of mass 2000 g covers a maximum vertical distance of 6 m when it is projected at an angle of 45° from the ground. Calculate the velocity with which it was thrown. Take g = 10 m/s^{2},
a) 12.10 m/s
b) 15.49 m/s
c) 2.155 m/s
d) 12.0 m/s
Ans: (a)
Solution:
Here, h = 6
θ = 45°
g = 10
h = (v sinθ)^{ 2}/2g is the formula for maximum height.
Solving the equation by substituting the values,
h= (v^{2}*[sin(45°)]^{2})/(2*10)
h = v^{2}X (1/√2 )^{2}/20
6 = [v^{2}X (1/2)]/20
6 X 20 X 2 = v^{2}
240 = v^{2}
v = 15.49
we get the initial velocity as 15.49 m/s.
Q.2. A body is projected with a velocity of 20ms^{−1} at 50° to the horizontal plane. Find the time of flight of the projectile.
Solution:
Initial Velocity V_{o} = 20ms^{−1}
And angle θ=50°
So, Sin 50° = 0.766
And g= 9.8
Now formula for time of flight is:
T = 2⋅u⋅sinθ.g
T = 2×20×sin50°×9.8
= 2×20×0.766×9.8
= 30.64×9.8
T = 3.126 sec
Therefore time of flight is 3.126 second.
Q.1. What is the concept of acceleration in horizontal and vertical projectile motion?
Ans: The only force acting on an object when it is projected in the air with some speed is the acceleration due to gravity (g). There is no acceleration in the horizontal direction because it acts vertically downwards, which means that the particle’s velocity remains constant in the horizontal direction.
Q.2. What is the trajectory?
Ans: A trajectory is known as the path followed by a projectile. When an object is thrown into space and upon which the only force acting is the force of gravity, then it is termed as a projectile. This doesn’t mean that other forces are not acting on it, but gravity is the primary force acting on a projectile. Due to this, the effects of other forces are minimised.
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