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Page 1 PROJECTILE MOTION & VECTORS Time of flight : T = g sin u 2 ? Horizontal range : R = g 2 sin u 2 ? Maximum height : H = g 2 sin u 2 2 ? Trajectory equation (equation of path) : y = x tan ? – ? 2 2 2 cos u 2 gx = x tan ? (1 – R x ) Projection on an inclined plane y x ? ? Up the Incline Down the Incline Range ? ? ? ? ? 2 2 cos g ) cos( sin u 2 ? ? ? ? ? 2 2 cos g ) cos( sin u 2 Time of flight ? ? cos g sin u 2 ? ? cos g sin u 2 Angle of projection with incline plane for maximum range 2 4 ? ? ? 2 4 ? ? ? Maximum Range ) sin 1 ( g u 2 ? ? ) sin 1 ( g u 2 ? ? RELATIVE MOTION B A AB v v ) B to respect with A of velocity ( v ? ? ? ? ? B A AB a a ) B to respect with A of on accelerati ( a ? ? ? ? ? Relative motion along straight line - A B BA x x x ? ? ? ? ? Page 2 PROJECTILE MOTION & VECTORS Time of flight : T = g sin u 2 ? Horizontal range : R = g 2 sin u 2 ? Maximum height : H = g 2 sin u 2 2 ? Trajectory equation (equation of path) : y = x tan ? – ? 2 2 2 cos u 2 gx = x tan ? (1 – R x ) Projection on an inclined plane y x ? ? Up the Incline Down the Incline Range ? ? ? ? ? 2 2 cos g ) cos( sin u 2 ? ? ? ? ? 2 2 cos g ) cos( sin u 2 Time of flight ? ? cos g sin u 2 ? ? cos g sin u 2 Angle of projection with incline plane for maximum range 2 4 ? ? ? 2 4 ? ? ? Maximum Range ) sin 1 ( g u 2 ? ? ) sin 1 ( g u 2 ? ? RELATIVE MOTION B A AB v v ) B to respect with A of velocity ( v ? ? ? ? ? B A AB a a ) B to respect with A of on accelerati ( a ? ? ? ? ? Relative motion along straight line - A B BA x x x ? ? ? ? ? CROSSING RIVER A boat or man in a river always moves in the direction of resultant velocity of velocity of boat (or man) and velocity of river flow. 1. Shortest Time : 2. Shortest Path : velocity along the river, v x = 0 and velocity perpendicular to river v y = 2 R 2 mR v v ? The net speed is given by v m = 2 R 2 mR v v ? at an angle of 90º with the river direction. velocity v y is used only to cross the river, Page 3 PROJECTILE MOTION & VECTORS Time of flight : T = g sin u 2 ? Horizontal range : R = g 2 sin u 2 ? Maximum height : H = g 2 sin u 2 2 ? Trajectory equation (equation of path) : y = x tan ? – ? 2 2 2 cos u 2 gx = x tan ? (1 – R x ) Projection on an inclined plane y x ? ? Up the Incline Down the Incline Range ? ? ? ? ? 2 2 cos g ) cos( sin u 2 ? ? ? ? ? 2 2 cos g ) cos( sin u 2 Time of flight ? ? cos g sin u 2 ? ? cos g sin u 2 Angle of projection with incline plane for maximum range 2 4 ? ? ? 2 4 ? ? ? Maximum Range ) sin 1 ( g u 2 ? ? ) sin 1 ( g u 2 ? ? RELATIVE MOTION B A AB v v ) B to respect with A of velocity ( v ? ? ? ? ? B A AB a a ) B to respect with A of on accelerati ( a ? ? ? ? ? Relative motion along straight line - A B BA x x x ? ? ? ? ? CROSSING RIVER A boat or man in a river always moves in the direction of resultant velocity of velocity of boat (or man) and velocity of river flow. 1. Shortest Time : 2. Shortest Path : velocity along the river, v x = 0 and velocity perpendicular to river v y = 2 R 2 mR v v ? The net speed is given by v m = 2 R 2 mR v v ? at an angle of 90º with the river direction. velocity v y is used only to cross the river, therefore time to cross the river, t = y v d = 2 R 2 mR v v d ? and velocity v x is zero, therefore, in this case the drift should be zero. ? v R – v mR sin ? = 0 or v R = v mR sin ? or ? = sin –1 ? ? ? ? ? ? ? ? mR R v v RAIN PROBLEMS Rm v ? = R v ? – m v ? or v Rm = 2 m 2 R v v ?Read More
FAQs on Important Formulas: Motion in a Plane
| 1. What are the main formulas I need to memorise for projectile motion in NEET Physics? |  |
Ans. Projectile motion uses key equations: horizontal displacement (x = ut cos θ), vertical displacement (y = ut sin θ - ½gt²), time of flight (T = 2u sin θ/g), and range (R = u² sin 2θ/g). Maximum height follows H = u² sin² θ/2g. These kinematic equations form the foundation for solving trajectory problems and are essential for NEET success.
| 2. How do I apply vector equations to solve motion in a plane problems quickly? | |
Ans. Vector equations decompose motion into components using position vector r = ut + ½at² and velocity vector v = u + at. Resolving forces and displacements into horizontal and vertical components simplifies calculations. This component method eliminates confusion and accelerates problem-solving during exams by treating perpendicular directions independently.
| 3. Why does uniform circular motion require centripetal acceleration when the speed stays constant? | |
Ans. Although speed remains constant in uniform circular motion, velocity continuously changes direction. Centripetal acceleration (a = v²/r or ω²r) is necessary to alter this direction toward the circle's centre. This concept clarifies why acceleration exists without speed change-it's directional, not magnitude-based.
| 4. What's the difference between angular velocity and linear velocity in circular motion formulas? | |
Ans. Angular velocity (ω) measures rotational rate in radians per second, whilst linear velocity (v) describes tangential speed. They relate through v = ωr, where r is radius. For NEET problems, using angular velocity simplifies calculations when dealing with rotating systems, while linear velocity suits trajectory-based questions in motion on inclined planes.
| 5. How do I choose between using displacement formulas and velocity formulas when solving 2D motion questions? | |
Ans. Use displacement formulas (s = ut + ½at²) when finding position or time; employ velocity formulas (v = u + at, v² = u² + 2as) when finding final speed or acceleration. For motion in a plane, identify what's given and what's required, then select kinematic equations accordingly to avoid unnecessary steps during NEET exams.
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