JEE Physics Motion In A Plane Notes Set 02

Projectile Motion

A projectile moves under gravity alone. Its horizontal velocity stays constant at \( u\cos\theta \), while the vertical motion has constant acceleration g downwards starting from \( u\sin\theta \). The two motions are independent and share only the time.

Standard Results for Level Ground

Angles \( \theta \) and \( 90^\circ - \theta \) give the same range. The range is maximum at 45°, where \( R = 4H \). At the top of the path the velocity is horizontal, perpendicular to g, and the speed is \( u\cos\theta \) (minimum). Because only the vertical component changes, the speed at any height depends only on the height, which gives the same landing speed for any direction of projection from the same point and speed.

Velocity at Any Instant

After time t: \( v_x = u\cos\theta \), \( v_y = u\sin\theta - gt \). The angle \( \alpha \) of the velocity with the horizontal satisfies \( \tan\alpha = \frac{v_y}{v_x} \) and the speed is \( v = u\cos\theta\sec\alpha \). The rate of change of speed is the component of g along the velocity, and the radius of curvature is \( r = \frac{v^2}{g\cos\alpha} \).

Projectile on an Inclined Plane

Resolve motion along and perpendicular to the plane. For a plane inclined at \( \beta \) and projection at angle \( \alpha \) to the plane, the acceleration components are \( g\sin\beta \) along the plane (down) and \( g\cos\beta \) perpendicular to it. The time of flight is \( T = \frac{2u\sin\alpha}{g\cos\beta} \) and the range along the plane is \( R = \frac{2u^2\sin\alpha\cos(\alpha+\beta)}{g\cos^2\beta} \).

Relative Velocity in Two Dimensions

\( \vec{v}_{AB} = \vec{v}_A - \vec{v}_B \). Two projectiles launched from the same point at the same time have zero relative acceleration, so each moves in a straight line relative to the other.

River-Boat and Rain-Man Problems

Let \( v_b \) be the boat's speed in still water and \( v_r \) the speed of the river. To cross in the shortest time, point the boat perpendicular to the bank: time = width / \( v_b \). To cross along the shortest path (straight across), aim upstream so that \( v_b\sin\phi = v_r \); the crossing speed is \( \sqrt{v_b^2 - v_r^2} \). For rain, the velocity of rain relative to a moving man is \( \vec{v}_{rain} - \vec{v}_{man} \), and an umbrella must be held along this relative velocity.

Solved Examples

Example 1. A ball is thrown at 20 m/s at 30° to the horizontal. Find the range (g = 10 m/s²).
\( R = \frac{400 \times \sin60^\circ}{10} = 20\sqrt{3} \approx 34.6 \) m.

Example 2. A stone is projected at 40 m/s at 30° with the horizontal. Find the maximum height.
\( H = \frac{1600 \times 0.25}{20} = 20 \) m.

Example 3. A boat with speed 10 km/h in still water crosses a 2 km wide river flowing at 6 km/h by the shortest path. Find the time.
Crossing speed = \( \sqrt{100 - 36} = 8 \) km/h, so the time = 2/8 h = 15 min.

Example 4. A body is projected horizontally at 20 m/s from a 45 m high cliff. How far from the base does it land?
Time \( = \sqrt{\frac{2 \times 45}{10}} = 3 \) s, so the horizontal distance = 20 × 3 = 60 m.

Key Points to Remember

1. The horizontal velocity is constant; only the vertical component changes.
2. At the top of the path the velocity is horizontal and the speed is minimum.
3. Time of flight, height and range all follow from the vertical and horizontal motions separately.
4. For the same speed and height, the landing speed does not depend on the direction of projection.
5. For relative motion subtract velocity vectors, and use components.