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Pls. explain equation of trajectory motion or give a suitable link.
Pls. explain equation of trajectory motion or give a suitable link.


2 years ago

Arun
25768 Points
							 A body projected into the space and is no longer being propelled by fuel is called a projectile.To analyze the projectile motion we use the following concept "Resolution of two dimensional motion into two one dimension motion" as discussed earlier. Hence it is easier to analyze the motion of projectile as composed of two simultaneous rectilinear motions which are independent of each other:(a) Along the vertical y-axis with a uniform downward acceleration 'g' and(b) Along the horizontal x-axis with a uniform velocity forward.Consider a particle projected with an initial velocity u at an angle θ with the horizontal x-axis as shown in figure shown below. Velocity and accelerations can be resolved into two components:                                                Velocity along x-axis = ux = u cos θAcceleration along x-axis ax = 0Velocity along y-axis = uy = u sin θAcceleration along y-axis ay = -gHere we use different equation of motions of one dimension derived earlier to get the different parameters.                           …... (a)            …... (b)         v2 = v02 – 2g (y-y0)                      …... (c)  Total Time of FlightWhen body returns to the same horizontal level, the resultant displacement in vertical y-direction is zero. Use equation b.Therefore, 0 = (u sin θ) t - (½)gt2,As t cannot equal to zero, then, total time of flight,Horizontal RangeHorizontal Range (OA=X)   = Horizontal velocity × Time of flight= u cos θ × 2 u sin θ/gSo horizontal range,Maximum HeightAt the highest point of the trajectory, vertical component of velocity is zero.Therefore, 0 = (u sin θ)2 - 2g HmaxSo, maximum height would be,  Equation of TrajectoryAssuming the point of projection as the origin of co-ordinates and horizontal direction as the x-axis and vertical direction as the y-axis. Let P (x, y) be the position of the particle at instant after t second.Then x = u cos θ.t and y = u sin θ.t - 1/2 gt2Eliminating 't' form the above equations, we get,y = x tan θ - (gx2/2u2cos2θ)This is the equation of trajectory which is a parabola (y = ax + bx2).

2 years ago
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• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions