Question

what is projectile motion and all the formulaes for trajectory motion horizontal range sent one of flight and descent ​

Answer 1

Explanation:

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Answer 2

$\mathcal{PROJECTILE\:\:MOTION}$

Projectile is the name given to a body thrown with some initial velocity at some angle with the horizontal direction, and then allowed to move in two dimensions under the action of gravity alone.

The path followed by a projectile is called its trajectory.

Some assumptions of projectile motion are as follows:

•There is no resistance due to air.

•The effect due to the curvature of earth is negligible.

•The effect due to the rotation of earth is negligible.

•For all points of the trajectory, the acceleration due to gravity g is constant in magnitude and direction.

Calculation of Various Parameters in Projectile Motion

Let a particle is projected with velocity u at an angle $\theta$ with the horizontal from point O. We can write the x and y components of the initial velocity as $\sf{u_x=ucos\theta}$

$\sf{u_y=usin\theta}$.

The acceleration of the particle in x and y direction is $\sf{a_x=0\:and\:a_y=-g}$

Time of Flight

For vertical upward motion, using $\sf{v_y=u_y+a_yt,}$ we get $\sf{u_y=usin\theta\:and\:a_y=-g}$

Hence, $\sf{0=u_y+a_yt=usin\theta-gt}$

$\implies\sf{t=\dfrac{u_y}{g}=\dfrac{usin\theta}{g}}$

Now the time taken to go up is equal to the time taken to come down, so time of flight

$\sf{T=2t=\dfrac{2u_y}{g}=\dfrac{2usin\theta}{g}}$

Maximum Height

Using $\sf{v_y^2=u_y^2+2a_ys\implies\:0=(usin\theta)^2-2gH}$, we get

$\sf{H=\dfrac{u^2sin^2\theta}{2g}}$

Horizontal Range

In horizontal direction, the acceleration of the particle is zero, horizontal component of the velocity is constant. Hence displacement in the horizontal direction can be written as

$\sf{x=R=u_x\times\:T}$

$\implies\:\sf{ucos\theta\:T=ucos\theta\times\:\left(\dfrac{2usin\theta}{g}\right)}$

$\sf{=\dfrac{u^2(2sin\theta\:cos\theta)}{g}}$

$\sf{R=\dfrac{u^2sin\theta}{g}}$

Equation of Trajectory

A projectile thrown with velocity u at an angle $\theta$ with the horizontal. The velocity u can be resolved into two rectangular components: $\sf{ucos\theta}$ component along X-axis and $\sf{usin\theta}$ component along Y-axis.

For horizontal motion, $\sf{x=ucos\theta\times\:t}$

$\implies\sf{t=\dfrac{x}{ucos\theta}.....(1)}$

For vertical motion, $\sf{y=(usin\theta)t-\dfrac{1}{2}gt^2.....(2)}$

From equation (1) and (2),

$\sf{y=usin\theta\left(\dfrac{x}{ucos\theta}\right)-\dfrac{1}{2}g\left(\dfrac{x^2}{u^2cos^2\theta}\right)}$

$\sf{y=xtan\theta-\dfrac{1}{2}\dfrac{gx^2}{u^2cos^2\theta}.....(3)}$

This equation shows that the trajectory of projectile is parabolic because it is similar to the equation of parabola.

It is known as the equation of trajectory. It is an equation of parabola. Hence, path of projectile is parabolic.

Equation (3) can also be written as:

$\sf{xtan\theta-\dfrac{x^2}{\dfrac{2u^2cos^2\theta}{g}}}$

$\sf{=xtan\theta-\dfrac{x^2\:sin\theta}{\left(\dfrac{2u^2cos\theta\:cos\theta}{g}\right)cos\theta}}$

$\sf{=xtan\theta-\dfrac{x^2tan\theta}{R}}$

$\sf{y=xtan\theta\left(1-\dfrac{x}{R}\right)}$