Name: Parameters of Motion and Trajectory of Projectile
Uploaded: 2019-07-19
Duration: 10 min 13 s
Description: Mathematics is a great tool to analyse any motion. The expressions for the height, range and time of flight of projectile motion can be derived using math. L...

Question

What is a projectile? Derive expression for the time of flight and horizontal range of a projectile.

Accepted Answer

When a particle is thrown at an angle with the earth’s surface, the particle follows a curved path before falling into the ground. This is called the projectile motion. The path that the particle follows is called the projectile and the motion of the particle is called the projectile motion.
In projectile motion, the velocity of the particle can be broken into its component along the x-axis and the y-axis. The velocity along the x-axis gives the horizontal motion of the particle and the velocity of the particle along y-axis gives the vertical motion of the particle.
When the particle moves in the projectile, the only acceleration acting on the particle is the acceleration due to gravity. The acceleration due to gravity acts on the particle in the vertical direction. In the horizontal direction, no acceleration acts on the particle.

Time of Flight

Let us consider the projectile motion of a projectile, to derive an expression for the time of flight of the projectile, as shown in the following figure

As clear from the figure, the initial position of the projectile is at O, when time, $t = 0$ . The projectile is launched from O with an initial velocity of $u$ , at an angle $θ$ , from the horizontal surface. Here, $θ$ is called the angle of elevation with respect to the x-axis or the horizontal surface. The projectile moves with its initial velocity till a certain height $h$ , away from the horizontal surface, from where it was thrown. Clearly, at a time $t$ , the final velocity of the projectile $v$ , on reaching the maximum height $h$ is equal to zero because the projectile stops here, before returning to the surface.

On separating the initial velocity $u$ into its horizontal as well as vertical components, we have

$u_{x} = u \cos θ u_{y} = u \sin θ$

where
$u$ is the initial velocity of the projectile
$u_{x}$ is the horizontal component of the initial velocity of the projectile
$u_{y}$ is the vertical component of the initial velocity of the projectile
$θ$ is the angle of elevation of the projectile with respect to the x-axis

Similarly, the acceleration of the projectile can also be separated into its horizontal as well as vertical components as given below.

$a_{x} = a \cos θ a_{y} = a \sin θ$

Clearly, $a_{x} = 0$ since the projectile is moving upwards

$a_{y} = - g = - 9.8 m / s^{2}$ since the acceleration of the projectile is acting against the acceleration due to gravity.

Now, let us consider the horizontal motion of the projectile from $t = 0$ to time $t$ when the projectile reaches its maximum height.

Using the first equation of motion we have

$v_{x} = u_{x} + a_{x} t = u_{x} + (0) t = u_{x}$ .............(1)

Using the second equation of motion, we have

$x = u_{x} t + \frac{1}{2} a_{x} t^{2} = u_{x} t + \frac{1}{2} (0) t^{2} = u_{x} t$ ...............(2)

Now, let us consider the vertical motion of the projectile from $t = 0$ to time $t$ when the projectile reaches its maximum height.

Using the first equation of motion we have

$v_{y} = u_{y} + a_{y} t = u \sin θ - g t$ .............(3)

Using the second equation of motion, we have

$y = u_{y} t + \frac{1}{2} a_{y} t^{2} = (u \sin θ) t - \frac{1}{2} g t^{2}$ ...............(4)

We know that at time $t$ , when the projectile is at a maximum height $h$ , the vertical component of the final velocity of the projectile is equal to zero because, at this instant, the projectile comes to a stop, before falling back to the ground. Clearly, equation (3) can be written as

$v_{y} = 0 \Rightarrow u \sin θ - g t = 0$

$\Rightarrow t = \frac{u \sin θ}{g}$ ..............(5)

Now, we know that time of flight of a projectile is equal to twice the time taken by the projectile to attain its maximum height. Therefore, the time of flight of a projectile is given by

$T = 2 t = \frac{2 u \sin θ}{g}$

where
$T$ is the time of flight of a projectile
$t$ is the time taken by the projectile to attain maximum height
$u$ is the initial velocity of the projectile
$θ$ is the angle of elevation of the projectile
$g$ is the acceleration due to gravity

Horizontal Range

Obviously, the horizontal range $R$ is the horizontal distance covered by the projectile with the uniform velocity $u \cos θ$ in a time equal to the time of flight.

$∴ R = u \cos θ \times T = u \cos θ \times \frac{2 u \sin θ}{g} \Rightarrow R = \frac{u^{2} (2 \sin θ \cos θ)}{g} A s 2 \sin θ \cos θ = \sin 2 θ \Rightarrow R = \frac{u^{2} \sin 2 θ}{g}$

What is a projectile? Derive expression for the time of flight and horizontal range of a projectile.

Important Questions on Projectile Motion

Learn and Prepare for any exam you want !