Projectile Motion

katie.barclay
Note by katie.barclay, updated more than 1 year ago
katie.barclay
Created by katie.barclay over 4 years ago
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Bachelor Degree Applied Maths (Motion) Note on Projectile Motion, created by katie.barclay on 06/22/2015.
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Step 1 - Construct Cartesian coordinates such that the y axis is pointing in the direction of the outward normal to the Earth's surface. Take the origin of the coordinate system to be the point of projection of P. Step 2 - With these coordinates in place. the position of the particle is x(t) = x(t)e1 + y(t)e2. The particle moves with constant acceleration g = ge1. The initial velocity of the particle is u = ucosa e1 + usina e2. Step 3 - The solution to the projectile motion in vector form is x'(t) = ucosa e1 + usina e2 - gt e2, x(t) = (-gt^2)/2 e2 + (ucosa e1 + usina e2)t.In component form the solution is x(t) = utcosa, y(t) = utsina - (gt^2)/2.Time of Flight - The time of flight is determined by solving the equation y(t) = 0 (finding the times the particle is in contact with the ground). This gives t = 0 or t = (2usina)/g. Therefore the time of flight is T = (2u/g)sina.Range - The range of the particle is the total horizontal distance covered during the flight (the value of x when y=0). So, R = (u^2 sin2a)/g. As sin2aMaximum Height - Maximum height occurs when y' = 0, when t = (u/g) sina. The maximum height is therefore (u^2 sin^2 a) / 2g.Particle Tragectory - The tragectory or path followed by the particle is obtained by eliminating t from the equations x(t) = utcosa, y(t) = utsina - (gt^2)/2. This gives y = xtana - gx^2 / 2u^2cos^2a. The resulting tragectory is part of an inverted parabola.

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