Question

Q35 0 What do you understand by the term proiectile explain with the help of an
example Show that the path of projectile projected horizontally from ac
ad horizontally from a certain height
from ground is a parabolic path.​

Answer 1

Projectile Motion:-

The body projected in the air with some velocity and then allowed to move through the air under the influence of the force of gravity is called projectile, and its motion is called projectile motion.

Features:-

• The time taken by the projectile to reach the point in the horizontal is called time of flight.

• The maximum height is attained by the projectile during its motion at half of its time of flight.

• The horizontal displacement of the projectile to reach the horizontal after projection is called the horizontal range of the projectile.

• The angle made by the projectile, or the initial velocity vector of the projectile, with the horizontal is called angle of projection.

• The point at which the projectile motion starts is called point of projection.

• The path of a projectile is called trajectory and it is always parabolic in shape (Proof is given below).

• The projectile motion is the resultant of horizontal motion with uniform velocity and vertical motion with uniform acceleration (i.e., -g).

The Parabolic Trajectory (Proof):-

Let,

$\longrightarrow\sf{u=}$ initial velocity

$\longrightarrow\sf{\theta=}$ angle of projection

$\longrightarrow\sf{t=}$ time of flight

If horizontal motion of the projectile is taken into consideration,

$\longrightarrow\sf{u_x=u\cos\theta}$

$\longrightarrow\sf{s_x=x}$

$\longrightarrow\sf{a_x=0}$

Then, by second kinematic equation,

$\longrightarrow\sf{s_x=u_xt+\dfrac {1}{2}a_xt^2}$

$\longrightarrow\sf{x=u\cos\theta t}$

$\longrightarrow\sf{t=\dfrac {x}{u\cos\theta}}\quad\quad\dots(1)$

If vertical motion of the projectile is taken into consideration,

$\longrightarrow\sf{u_y=u\sin\theta}$

$\longrightarrow\sf{s_y=y}$

$\longrightarrow\sf{a_y=-g}$

Then, by second kinematic equation,

$\longrightarrow\sf{s_y=u_yt+\dfrac {1}{2}a_yt^2}$

$\longrightarrow\sf{y=u\sin\theta t-\dfrac {1}{2}gt^2}$

$\longrightarrow\sf{y=\dfrac {u\sin\theta x}{u\cos\theta}-\dfrac {gx^2}{2u^2\cos^2\theta}}$

$\longrightarrow\sf{y=x\tan\theta-\dfrac {g}{2u^2\cos^2\theta}x^2}$

Since $\sf{\tan\theta}$ and $\sf{\dfrac {g}{2u^2\cos^2\theta}}$ are constants,

$\longrightarrow\sf{y\propto x-x^2\quad\quad\dots(2)}$

The relation (2) shows that the trajectory is a downward parabola.

Hence the Proof!