JEE Physics Motion In A Plane Notes Set 01

Projectile Motion

A projectile moves under gravity alone. The horizontal velocity stays constant and the vertical motion is free fall, and the two motions are independent. For a launch at speed u and angle \( \theta \): \( u_x=u\cos\theta \) and \( u_y=u\sin\theta \).
The path is a parabola \( y=x\tan\theta-\frac{gx^2}{2u^2\cos^2\theta} \). At the highest point the velocity is horizontal and the acceleration is still g downwards.

Time of Flight, Height and Range (level ground)

\( T=\frac{2u\sin\theta}g \), \( H=\frac{u^2\sin^2\theta}{2g} \), \( R=\frac{u^2\sin2\theta}g \).
Range is greatest at \( 45^\circ \), where \( R_{max}=\frac{u^2}g \) and \( H=\frac R4 \). Angles \( \theta \) and \( 90^\circ-\theta \) give the same range. Also \( \tan\theta=\frac{4H}R \) and \( T^2=\frac{8H}g \).

Horizontal Projection from a Height

For a body thrown horizontally with speed u from height h: time of fall \( T=\sqrt{\frac{2h}g} \), range \( x=uT \) and speed on landing \( \sqrt{u^2+2gh} \).
At time t the velocity components are \( u \) and \( gt \), so the angle with the horizontal is given by \( \tan\alpha=\frac{gt}u \).

Oblique Projection from a Height

Take the point of projection as the origin and write the vertical displacement as minus h at the end. The time of flight is the positive root of \( -h=u\sin\theta\,t-\frac12gt^2 \).
A projectile launched from a moving carrier has the carrier's velocity added to its own, so a ball thrown straight up from a moving ship travels a parabola for an observer on the shore.

Relative Motion in a Plane

Relative velocity is a vector difference: \( \vec v_{AB}=\vec v_A-\vec v_B \). Two bodies with constant velocities collide only if their relative velocity points along the line joining them.
The closest approach occurs when the relative position is perpendicular to the relative velocity. Two projectiles launched together see each other moving in a straight line because gravity cancels in the relative acceleration.

Rain and Man Problem

Rain with velocity \( \vec v_R \) appears to a man moving with \( \vec v_M \) to have velocity \( \vec v_{RM}=\vec v_R-\vec v_M \). For vertical rain and a man walking at speed v, the umbrella must be tilted from the vertical by \( \theta \) with \( \tan\theta=\frac{v}{v_R} \).
If the man moves faster, the apparent rain comes more slanted towards him.

River and Boat Problem

Let the boat speed in still water be v, the river speed u and the width d. The velocity of the boat relative to the ground is the vector sum of v and u.
Minimum time: head perpendicular to the banks, time \( \frac dv \) (independent of u), drift \( \frac{ud}v \). Shortest path (needs \( v>u \)): head upstream at angle \( \theta \) from the perpendicular with \( \sin\theta=\frac uv \), time \( \frac d{\sqrt{v^2-u^2}} \).
Going a distance L upstream and back takes \( \frac L{v-u}+\frac L{v+u} \).

Pursuit Problems

In a regular n-sided polygon where each person always runs towards the next at speed u, the rate of approach along the line to the neighbour is \( u\left(1-\cos\frac{2\pi}n\right) \), so they meet after \( \frac{d}{u(1-\cos(2\pi/n))} \).
For four persons on a square the time is \( \frac du \) and for six persons on a hexagon it is \( \frac{2d}u \).