Motion in Two Dimension |

 Introduction

Important definitions

$\to$ Projectile: A body thrown into space and allowed to move only under the action of gravity is called projectile.

$\to$ Projectile Motion: The motion shown by projectile is called projectile motion.

$\to$ Angle of Projection: The angle with the horizontal at which the body is projected is called the angle of projection.

$\to$ Velocity of Projection: The velocity with which the body is thrown is called the velocity of projection.

$\to$ Point of Projection: The point from which the body is projected in the air is called a point of projection.

$\to$ Trajectory of Projectile: The path followed by a projectile in the air is called the trajectory of the projectile.

$\to$ Range of Projection: The horizontal distance travelled by the body performing projectile motion is called the range of the projectile.

Assumptions of Projectile Motion

$\to$ The free-fall acceleration is constant over the range of motion and it is directed downward.

$\to$It is reasonable as long as the range is small compared to the radius of the Earth.

$\to$ The effect of air friction is negligible.

The projectile motion near the surface of the earth (gravity is constant) consists of two independent motions.$\\$

$\to$A horizontal motion at a constant speed.$\\$

$\to$A vertical motion subjected to a constant acceleration due to gravity.$\\$

$\textbf {Note:}$ In projectile motion acceleration is constant but velocity is changing in magnitude and direction at every instant.

Oblique Projection

When a body is thrown into space at an angle θ with horizontal with initial velocity u from the ground. This is called oblique projection.

Few Deductions:

$\to$ In X axis initial velocity, $u_x = u \cos\theta$ acceleration, $a_x = 0$ i.e the motion is uniform.

$\to$ In Y axis initial velocity, $u_y=u \sin \theta$ acceleration $a_y=- g$ i.e the motion is uniformly accelerated.

$\to$ Horizontal distance travelled $(x)=u_x t + \dfrac{1}{2} a_xt^2= u t \cos \theta$ (Where $u_x=u \cos \theta$)

$\to$ Vertical distance covered $(y)= u_yt + \dfrac{1}{2} a_yt^2 = u t \sin \theta -\dfrac{1}{2}gt^2$

$\to$ Instantaneous Velocity (v):

Magnitude: $\sqrt{v_x^2 + v_y^2}$

Direction: $\tan \alpha = \dfrac{v_y}{v_x}$

$\to$ Equation of trajectory: $y=x \tan θ - \dfrac{gx^2}{2u^2cos^2\theta}$. This is the equation of trajectory in projectile motion

$\to$ $y \propto x^2$ i.e. The projectile motion is always parabolic in nature.

Important Terms in Projectile:

$\to$ Time of flight (T) = $\dfrac {2u_y}{g}=\dfrac{2u \sin \theta}{g} = \sqrt{\dfrac{2H_{max}}{g}} = \dfrac{R}{u \cos \theta}$

$\to$ Horizontal Range(R)=$u_x T = \dfrac{2u_xu_y}{g}=\dfrac{u^2 \sin 2 \theta }{g}=4H_{\text{max}} \cot \theta$

$\to$ Maximum Height ($H_{max})=\dfrac{{u_y}^2}{2g} = \dfrac{u^2\sin^2\theta}{2g} = \dfrac{gT^2}{8} = \dfrac{1}{4} R \tan \theta$

For a projectile to obtain same range the angles of projection should be $\theta$ and $90- \theta$. (In a condition that speed is constant).

For maximum horizontal range, the angle of projection should be 45 degrees.

In projectile motion, the horizontal component of velocity $(u \cos \theta)$, acceleration (g) and mechanical energy remains constant while speed, velocity, the vertical component of velocity $(u \sin \theta)$, momentum kinetic energy and potential energy all change. Velocity and KE are maximum at the point of projection while minimum (but not zero) at the highest point.

Angle between acceleration and velocity gradually decreases from $90+\theta$ to $90- \theta$ while acceleration and velocity are right angles at the highest point.

At the highest point,$\\$

$\to$ Velocity(V) $= u\cos\theta \\$

$\to$ Momentum$=mu\cos\theta = p\cos\theta \\$

$\to$ Kinetic Energy$=E\cos^2\theta$

(Substitute $v=u\cos\theta$ in general eqn of K. E)$\\$

$\to$ Potential Energy$=E\sin^2 \theta$

From point of projection to maximum height$\\$

$\to$ Change in direction $=\theta$$\\$

$\to$ Change in velocity $= u\sin\theta \\$

$\to$ Change in momentum $= mu\sin \theta \\$

$\to$ Decrease in K.E(increase in P.E) $= E\sin^2\theta \\$

$\to$ Change in speed $= u-ucos\theta =2u\sin^2\dfrac{\theta }{2} \\$$\\$

$\to$ average velocity = $\dfrac{u}{2} \sqrt{1+ 3 \cos^2 \theta}$

From point of Projection to ground$\\$

$\to$ Change in speed =0$\\$

$\to$ Change in velocity $= 2u\sin\theta$$\\$

$\to$ Change in K.E and P.E = zero

$\to$ Change in momentum $= 2P\sin\theta$$\\$

$\to$ Change in direction of motion $= 2\theta$

$\to$ average velocity = $u \cos \theta$

Projection from Height

Suppose a body is thrown horizontally from a point O with velocity u. Point O is at height h above the ground.

Few Deductions:

$\to$ In X axis initial velocity, $u_x = u$ acceleration, $a_x = 0$ i.e the motion is uniform.

$\to$ In Y axis initial velocity, $u_y=0$ acceleration $a_y= g$ i.e the motion is uniformly accelerated.

$\to$ Horizontal distance travelled $(x)=u_x t + \dfrac{1}{2} a_xt^2= u t$ (Where $u_x=u$)

$\to$ Vertical distance covered $(y)= u_yt + \dfrac{1}{2} a_yt^2 = \dfrac{1}{2}gt^2$

$\to$ Instantaneous Velocity (v):

Magnitude: $\sqrt{v_x^2 + v_y^2} = \sqrt{u^2 + g^2t^2}$

Direction: $\tan \beta = \dfrac{v_y}{v_x} = \dfrac{gt}{u}$

$\to$ Equation of trajectory: $y=h - \dfrac{gx^2}{2u^2}$. This is the equation of trajectory in projectile motion

$\to$ $y \propto x^2$ i.e. The projectile motion is always parabolic in nature.

Important Terms in Projectile:

$\to$ Time of flight (T) = $\sqrt{\dfrac{2h}{g}}$

$\to$ Horizontal Range(R)=$u T = u \sqrt{\dfrac{2h}{g}}$

$\to$ Final velocity (v):

Magnitude: $\sqrt{u^2 + g^2T^2} = \sqrt{u^2 + 2gh}$

Direction: $\theta = \dfrac{\sqrt{2gh}}{u}$

Better You Know

If a body is released from a plane moving horizontally with speed u at height h, its initial vertical component of velocity is zero and initial horizontal component of velocity is equal to that of the moving plane.

The motion of a projectile relative to another projectile is a straight line.

A body feels weightless in a projectile motion.

The linear momentum at the highest point is $mu \cos \theta$ and the kinetic energy is $\dfrac{1}{2} m u \cos\theta$(Oblique Projectile)

If a man aims to hit a target, he should point his gun in a direction higher than the target.

If a hunter aims monkey, and at that instant monkey falls then a bullet will hit the monkey.

If a large number of bullets are fired in all directions with speed v, then area bullets will spread is $\frac{\pi v^4}{g^2}$$\\$

(Area $=\pi r^2$, and $R_{max}=\frac{v^2}{g}$)

When a ball rolls off the top of the stairway with a horizontal velocity u having steps h m high and b m wide the ball will hit the edge of nth step given by $n=\frac {2hu^2}{gb^2}$.$\\$

If a boy can throw a ball to a maximum distance of R, then the maximum vertical height he can throw is $\dfrac{R}{2}$ But, The maximum vertical height reached is $\dfrac{R}{4}$.

The angle of projection at which horizontal range and maximum height are equal is $\tan^{-1} 4$ which is $75.96^{\circ}$.

In a javelin throw, if a player throws with an angle slightly less than 45 degree , he increase his probability of winning.

Time of flight will be highest when height attained by projectile is largest.

If air resistance is taken into account then Time of descent>Time of ascent

If a ball A is dropped from top of tower and another ball B is thrown horizontally from same height at same moment with same velocity, the ball B has greater speed when it reaches to ground.

When a body is projected at two angles $\theta$ and $90- \theta$ with horizontal(same speed), then$\\$

$\to$ $\dfrac{R_1}{R_2}=1:1$$\\$

$\to$ $\dfrac{H_1}{H_2} = \tan^2\theta \\$

$\to$ $\dfrac{T_1}{T_2} = \tan\theta \\$

$\to$ $R_1=R_2=R=4\sqrt{H_1H_2} \\$

If two bodies are projected horizontally from the top of the tower in opposite direction with velocities $u_1$ and $u_2$, then

$\to$ The time after which their velocities are perpendicular is $t=\dfrac {\sqrt{u_1u_2}}{g} \\$

$\to$ The time after which their position vectors are perpendicular is $t=\dfrac {2\sqrt{u_1u_2}}{g} \\$

$\to$ The distance between them at which their position vectors are perpendicular is $s=(u_1+u_2)\dfrac {\sqrt{u_1u_2}}{g} \\$

If a passenger drops a coin from a bus in a uniform motion, the path of the coin will appear

$\to$ A straight line to the passenger.

$\to$ A parabola to a person in ground.

If a ball A is dropped and another ball B is projected horizontally from same height h, they hit the ground simultaneously after time $\dfrac{2h}{g}$ But, the ball B will hit the ground with greater speed than ball A.

When air resistance is neglected:

$\to$ Maximum height, horizontal range, speed, momentum, final speed of the projectile at which are less than those when air resistance is neglected.

$\to$ Time of flight and the angle with which projectile hits the ground will be more than those when the air resistance is neglected.