JEE Physics Motion In A Plane Notes Set 06

Motion in a Plane: Projectile and Relative Motion

A projectile moves under gravity alone. Its horizontal velocity \( u\cos\theta \) never changes, while the vertical velocity changes as \( v_y=u\sin\theta-gt \). The acceleration is g downward at every point, including the top, where only the vertical velocity is zero. The speed is least at the top and the path is a parabola.

Projectile Formulas (level ground)

The range is the same for angles \( \theta \) and \( 90^\circ-\theta \) and is greatest at \( 45^\circ \), where \( R_{max}=\frac{u^2}{g}=4H \). For complementary angles the heights and flight times satisfy \( R=4\sqrt{H_1H_2} \) and \( t_1t_2=\frac{2R}{g} \). The highest point reachable by throwing straight up is half the maximum range.

For a body thrown horizontally with speed u from a height h, the flight time is \( \sqrt{\frac{2h}{g}} \), the range is \( u\sqrt{\frac{2h}{g}} \), and at time t the velocity makes an angle \( \tan^{-1}\frac{gt}{u} \) below the horizontal.

Relative Velocity

The velocity of A relative to B is \( \vec v_{AB}=\vec v_A-\vec v_B \). In problems with an observer in motion, always write the true (ground) velocity as relative velocity plus the observer's velocity. Two bodies on collision course must have the relative velocity directed along the line joining them. For minimum separation, the separation vector is perpendicular to the relative velocity.

Rain and Umbrella

If rain falls vertically at speed \( v_r \) and a man walks at \( v_m \), he feels rain with speed \( \sqrt{v_r^2+v_m^2} \), tilted forward from the vertical by \( \tan^{-1}\frac{v_m}{v_r} \). The umbrella is held along this relative velocity.

Boat in a River

Let the boat speed in still water be v and the river speed u. Going downstream the ground speed is \( v+u \) and upstream \( v-u \). To cross in minimum time, point the boat straight across: \( t_{min}=\frac dv \) and the drift is \( u\,t_{min} \). To reach the point directly opposite (possible only if \( v>u \)), steer upstream at angle \( \theta \) from the perpendicular with \( \sin\theta=\frac uv \); the crossing time is \( \frac{d}{v\cos\theta} \). If \( u>v \), the minimum drift is obtained with \( \sin\theta=\frac vu \) measured from the bank direction.

Chasing Problems

When n bodies at the corners of a regular polygon each move at speed v towards the next, they spiral to the centre. Look at two neighbours: the chaser closes in at v, while the body ahead moves at v along the next side, which is turned by the exterior angle \( \frac{2\pi}{n} \) from the line joining them. So the rate of approach is \( v\left(1-\cos\frac{2\pi}{n}\right) \) and the meeting time is the side divided by this rate. This gives \( \frac{2d}{3v} \) for an equilateral triangle, \( \frac dv \) for a square and \( \frac{2d}{v} \) for a regular hexagon.