स्थिति सदिश (Position Vector)|एकसमान वेग से द्विविमीय गति|वेग समीकरण (Velocity Equation)|विस्थापन समीकरण (Displacement|प्रक्षेप्य गति (Projectile Motion)|क्षैतिज प्रक्षेप्य (Horizontal Projectile)|तिर्यक प्रक्षेप्य (Oblique Projectile)|प्रक्षेप्य का उड्डयन काल (Flight Time of Projectile)|क्षैतिज परास (Horizontal Range)|अधिकतम ऊंचाई (Maximum Height)

What Is Projectile? (प्रक्षेप्य क्या है?)|Maximum Height (अधिकतम ऊंचाई)|Time Of Flight (उडडयन काल)|Range (क्षैतिज पराश)|OMR

Revision| प्रक्षेप्य गति|उड्डयन काल (Time Of Flight T)|प्राप्त की गई अधिकतम ऊंचाई|क्षैतिज परास (Horizontal Range Or Range "R")|अधिकतम क्षैतिज परास (Horizontal Range Or Range "R")|प्रक्षेप पथ की समीकरण|OMR|Summary

Revision| प्रक्षेप्य गति|उड्डयन काल (Time Of Flight T)|प्राप्त की गई अधिकतम ऊंचाई|क्षैतिज परास (Horizontal Range Or Range "R")|अधिकतम क्षैतिज परास (Horizontal Range Or Range "R")|प्रक्षेप पथ की समीकरण|OMR|Summary

Find the equation of trajectory, time of flight , maximum height and horizontal range of a projectile when projected at an angle theta with the vertical direction .

A particle is projected from origin in xy-plane and its equation ,of trajectory is given by y = ax - bx^(2) . The only acceleration ,in the motion is' f' which is constant and in -ve direction of, y-axis. (a) Find the velocity of projection and the angle of projection. (b) Point of projection is considered as origin and x-axis along the horizontal ground. Find the horizontal range: and maximum height of projectile. Projectile completes its flight in horizonal plane of projection.

Given that u_(x) = horizontal component of initial velocity of a projectile, u_(y) = vertical component of initial velocity, R = horizontal range, T = time of flight and H = maximum height of projectile. Now match the following two columns.

Assertion : Two projectile have maximum heights 4H and H respectively. The ratio of their horizontal components of velocities should be 1:2 for their horizontal ranges to be same. Reason : Horizontal range = horizontal component of velocity xx time of flight.

At a height of 45 m from ground velocity of a projectile is, v = (30hati + 40hatj) m//s Find initial velocity, time of flight, maximum height and horizontal range of this projectile. Here hati and hatj are the unit vectors in horizontal and vertical directions.

Projectile motion is a combination of two one-dimensional motion: one in horizontal and other in vertical direction. Motion in 2D means in a plane. Necessary condition for 2D motion is that the velocity vector is coplanar to the acceleration vector. In case of projectile motion, the angle between velocity and acceleration will be 0^@ltthetalt180^@ . During the projectile motion, the horizontal component of velocity ramains unchanged but the vertical component of velocity is time dependent. Now answer the following questions: A body is projected at angle of 30^@ and 60^@ with the same velocity. Their horizontal ranges are R_1 and R_2 and maximum heights are H_1 and H_2 , respectively, then A body is projected at angle of 30^@ and 60^@ with the same velocity. Their horizontal ranges are R_1 and R_2 and maximum heights are H_1 and H_2 , respectively, then

If T is the total time of flight, h is the maximum height and R is the range for horizontal motion, the x and y coordinates of projectile motion and time t are related as