Projectile Motion. Engineering Mechanics: Dynamics. D. Dane Quinn, PhD. Copyright c 2016 All rights reserved

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University of Akron Akron OH 44325 393 USA Copyright c 216

1 Engineering Mechanics: Dynamics D. Dane Quinn, PhD Department of Mechanical Engineering The University of Akron Akron OH USA Copyright c 216 All rights reserved D. D. Quinn (The University of Akron) c 216, D. Dane Quinn 1 / 8

2 Dynamics Describes the motion of a particle subject only to the force due to gravity. The trajectory (path through space) depends only on the initial conditions (position and velocity) of the object. g v B () B O rb/o() Coordinates and Directions F grav Only the gravitational force acts on the particle, while the motion of the particle is in the plane, so that we define x and y to describe the displacement of the ball as rb/o(t) = x(t)î+y(t)ĵ. y O rb/o x B ĵ î D. D. Quinn (The University of Akron) c 216, D. Dane Quinn 2 / 8

3 Dynamics Kinematics/Free Body Diagram With this, the acceleration of the ball is ā B = ẍî+ÿĵ, and the free body diagram is shown to the right. Equations of Motion mgĵ Finally, applying linear momentum balance to this particle F = mgĵ = m (ẍî+ÿĵ) = māb, so that [ ] [ ] = mẍ î+ mÿ +mg ĵ. Taking components in the î and ĵ directions provides î direction ẍ =, ĵ direction ÿ = g, (constant acceleration), D. D. Quinn (The University of Akron) c 216, D. Dane Quinn 3 / 8

4 Dynamics Integrating to find the velocity t { } ẍ(τ) = dτ ẋ(t) ẋ() =, (ẋ() v x ), t { } ÿ(τ) = g and the position } {ẋ(τ) = v x t t {ẏ(τ) = gτ +v y } ẋ(t) = v x, dτ ẏ(t) ẏ() = gt, (ẏ() v y ), ẏ(t) = gt+v y, dτ x(t) x() = v x t, (x() x ) x(t) = v x t+x, dτ y(t) y() = gt2 2 +v yt, (y() y ) y(t) = gt2 2 +v yt+y, These are valid for any particle undergoing projectile motion D. D. Quinn (The University of Akron) c 216, D. Dane Quinn 4 / 8

5 Trajectory y y y f y f x f x t f t t f x x f (x,y ) = (,) x(t) = v x t, t = x v x y(t) = gt2 2 +v yt, y(x) = gx2 2v 2 x + v y v x x. t D. D. Quinn (The University of Akron) c 216, D. Dane Quinn 5 / 8

6 Initial Velocity The initial velocity v B () can be represented as v B () = v x î+v y ĵ = v (C θ î+s θ ĵ), where v is the initial speed and θ is the initial angle of inclination. Therefore ẋ() v x = v C θ, ẏ() v y = v S θ. v θ ĵ î In terms of the initial speed and inclination, with (x,y ) = (,) ẋ(t) = v C θ, ẏ(t) = gt+v S θ, x(t) = v C θ t, y(t) = gt2 2 +v S θ t, while the trajectory becomes y(x) = gx2 2v 2 x + v y v x x = gx2 2v 2 C2 θ +T θ x. D. D. Quinn (The University of Akron) c 216, D. Dane Quinn 6 / 8

7 Auxiliary Conditions The motion can also be defined in terms of auxiliary conditions, rather than the initial conditions. Auxiliary conditions provide equations that can be used to solve for the inital conditions Secondary Position If the trajectory passes through a secondary point (x s,y s ) at some time t s, then x(t s ) = x s = v x t s +x, y(t s ) = y s = gt s 2 +v y t s +y 2, y s = g(x s x ) 2 2v 2 x + v y v x (x s x )+y. (x,y ) (x s,y s ) D. D. Quinn (The University of Akron) c 216, D. Dane Quinn 7 / 8

8 Auxiliary Conditions Maximum Height If the maximum heighty = y occurs at timet = t (with x =, and y = ) so that ẏ(t ) = = gt +v y, t = v y g, and y = y(t ) = v2 y 2g = v2 C2 θ 2g, y x(t ) = v x t = v xv y g = v2 S θ C θ. g D. D. Quinn (The University of Akron) c 216, D. Dane Quinn 8 / 8

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