18_KINE426_Projectile Motion

1 Andrew Nordin, PhD Assistant Professor, Department of Health & Kinesiology KINE 426 Exercise Biomechanics Linear Kinematics Chapter 8 In the figure on the left, the gymnast needs to travel through the air to the higher of the two uneven bars. Once airborne, imagine only the gymnast’s center of mass travelling from the low bar to the high bar . That is, a trajectory in the sagittal plane from the center of the low bar to the center of the high bar. If the high bar is 0.8m above the low bar, and the diagonal distance between the bars is 1.5m, what range needs to be covered between the high and low bars? Once airborne, the gymnast reached peak height in 0.48s. What was the gymnast’s peak height above the high bar ? During flight, what is the gymnast’s horizontal velocity ? What resultant velocity at takeoff produced this trajectory? When reaching the high bar, what is the gymnast’s vertical velocity ?

2 Projectile Motion ࠵? ! " = ࠵? # " + 2࠵?࠵? ࠵? ! = ࠵? #$ ࠵?࠵? ࠵? = ࠵? # ࠵? + % " ࠵?࠵? " d y a y = -9.81m/s 2 a x = 0 R θ v x v y v y = 0 R R d up d down d up d down t up t down t up t down t up = t down = 1/2 t flight t flight = t up + t down + projection height - projection height A B C Projectile Motion • Solving projectile motion problems – Initial conditions: identify known values – Resolve into components (vectors) • Separate vertical & horizontal components • Only flight time t flight used between axes – Known values during flight • a y = gravity = -9.81m/s 2 • a x = 0 (neglecting air resistance) • v y = 0 @ apex (d y ) – If projection height = 0 • t up = t down = ½ t flight & d y-up = d y-down – If projection height ≠ 0 • t up ≠ t down & d up ≠ d down • Separate upward & downward segments Uniform acceleration equations ࠵? ’ ( = ࠵? ) ( + 2࠵?࠵? ࠵? ’ = ࠵? )* ࠵?࠵? ࠵? = ࠵? ) ࠵? + + ( ࠵?࠵? (

4 Once airborne, the gymnast reached peak height in 0.48s. What was the gymnast’s peak height above the high bar ? ࠵? ! = ࠵? " + ࠵?࠵? ࠵? = ࠵? " ࠵? + 0.5࠵?࠵? # ࠵? ! # = ࠵? " # + 2࠵?࠵? d y = 0.8m 1.5m d x = 1.27m v y = 0m/s a y = -9.8m/s 2 a x = 0m/s 2 v x = d x / t flight x - horizontal v x = v R cos θ y - vertical up down v y = v R sin θ d up v up = 4.70m/s t up = 0.48s d down v down t down ࠵? $ = ࠵? "$ ࠵? + 0.5࠵? $ ࠵? # = (4.70࠵?/࠵?)(0.48࠵?) + 0.5(−9.8࠵?/࠵? # )(0.48࠵?) # = 2.268࠵? − 1.130࠵? = 1.128࠵? ࠵? !$ = ࠵? "$ + ࠵? $ ࠵? 0 = ࠵? "$ + (9.81࠵?/࠵? # )(0.48࠵?) ࠵? "$ = 4.70࠵?/࠵? Once airborne, the gymnast reached peak height in 0.48s. What was the gymnast’s peak height above the high bar ? ࠵? ! = ࠵? " + ࠵?࠵? ࠵? = ࠵? " ࠵? + 0.5࠵?࠵? # ࠵? ! # = ࠵? " # + 2࠵?࠵? d y = 0.8m 1.5m d x = 1.27m v y = 0m/s a y = -9.8m/s 2 a x = 0m/s 2 v x = d x / t flight x - horizontal v x = v R cos θ y - vertical up down v y = v R sin θ d up = 1.13m v up = 4.70m/s t up = 0.48s d down = 0.33m v down t down = (4.70࠵?/࠵?)(0.48࠵?) + 0.5(−9.8࠵?/࠵? # )(0.48࠵?) # = 2.268࠵? − 1.130࠵? = 1.13࠵? − ࠵?. ࠵?࠵? = 0.33࠵? ࠵?࠵?࠵?࠵?࠵? ℎ࠵?࠵?ℎ ࠵?࠵?࠵? 0 = ࠵? " + (9.81࠵?/࠵? # )(0.48࠵?) ࠵? "$ = 4.70࠵?/࠵? ࠵? !$ = ࠵? "$ + ࠵? $ ࠵? ࠵? $ = ࠵? "$ ࠵? + 0.5࠵? $ ࠵? #

5 During flight, what is the gymnast’s horizontal velocity ? ࠵? ! = ࠵? " + ࠵?࠵? ࠵? = ࠵? " ࠵? + 0.5࠵?࠵? # ࠵? ! # = ࠵? " # + 2࠵?࠵? d y = 0.8m 1.5m d x = 1.27m v y = 0m/s a y = -9.8m/s 2 a x = 0m/s 2 v x = d x / t flight x - horizontal v x = v R cos θ y - vertical up down v y = v R sin θ d up = 1.13m v up = 4.70m/s t up = 0.48s d down = 0.33m v down t down ࠵? $ = ࠵? "$ ࠵? + 0.5࠵? $ ࠵? # (0 − 0.33࠵?) = 0࠵? + 0.5(−9.81࠵?/࠵? # )࠵? # ∆࠵? = ࠵? ! − ࠵? " −0.33࠵? = 0.5(−9.81࠵?/࠵? # )࠵? # ࠵? %&’( = −0.33࠵? −4.91࠵?/࠵? # = 0.26࠵? ࠵? !)"*+, = ࠵? -. + ࠵? %&’( = 0.48࠵? + 0.26࠵? = 0.74࠵? During flight, what is the gymnast’s horizontal velocity ? ࠵? ! = ࠵? " + ࠵?࠵? ࠵? = ࠵? " ࠵? + 0.5࠵?࠵? # ࠵? ! # = ࠵? " # + 2࠵?࠵? d y = 0.8m 1.5m d x = 1.27m v y = 0m/s a y = -9.8m/s 2 a x = 0m/s 2 v x = d x / t flight x - horizontal v x = v R cos θ y - vertical up down v y = v R sin θ d up = 1.13m v up = 4.70m/s t up = 0.48s d down = 0.33m v down t down = 0.26s ∆࠵? = ࠵? ! − ࠵? " t flight = 0.74s ࠵? / = ࠵? / ࠵? !)"*+, ࠵? / = 1.27࠵? 0.74࠵? ࠵? / = 1.72࠵?/࠵?

7 Projectile Motion ࠵? ! " = ࠵? # " + 2࠵?࠵? ࠵? ! = ࠵? #$ ࠵?࠵? ࠵? = ࠵? # ࠵? + % " ࠵?࠵? " d y a y = -9.81m/s 2 a x = 0 R θ v x v y v y = 0 R R d up d down d up d down t up t down t up t down t up = t down = 1/2 t flight t flight = t up + t down + projection height - projection height A B C Projectile Motion • Solving projectile motion problems – Initial conditions: identify known values – Resolve into components (vectors) • Separate vertical & horizontal components • Only flight time t flight used between axes – Known values during flight • a y = gravity = -9.81m/s 2 • a x = 0 (neglecting air resistance) • v y = 0 @ apex (d y ) – If projection height = 0 • t up = t down = ½ t flight & d y-up = d y-down – If projection height ≠ 0 • t up ≠ t down & d up ≠ d down • Separate upward & downward segments Uniform acceleration equations ࠵? ’ ( = ࠵? ) ( + 2࠵?࠵? ࠵? ’ = ࠵? )* ࠵?࠵? ࠵? = ࠵? ) ࠵? + + ( ࠵?࠵? (

8 Summary • Kinematics • Spatial & temporal components of motion • Ignores forces causing motion • Motion capture/videography • Linear kinematics – Rectilinear – Curvilinear • Reference systems – Cartesian coordinate system (2D & 3D) – Global vs. Local coordinate systems y x (0,0) (x,0) (0,y) (x,y) Summary • Scalar vs. vector – Magnitude vs. magnitude & direction – Vector operations – Distance vs. displacement – Speed vs. velocity – Acceleration • Projectile motion – Parabola – Factors influencing projectiles – Uniform acceleration equations • Resolve into components (vertical vs. horizontal) d y v x v y a y = gravity a x = 0 R θ v x v y v R Displacement Distance