16_KINE426_Projectile Motion

3 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 y_up 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 v yf_up = 0 = v yi_down v yf_up = 0 = v yi_up ∆ d y_down d yi_up d yf_down = 0 t up t down d yi_up = 0 v yi_up v yf_down -v y d yf_down ∆ d y_up ∆ d y_down v yi_up v yf_down i vs. f x vs. y up vs. down initial vs. final horizontal vs. vertical vertical trajectory v x v x v x Variable subscripts Projectile Motion Shot putter releases shot @ 40° angle from a height of 2.2m, with a velocity of 13.3m/s How far will the shot travel? 40 o 2.2m 13.3m/s

6 Projectile Motion • Use uniform acceleration equations to solve for time to apex ( t up ) • Solve for apex height – Relative to takeoff height ࠵? " = ࠵? ",% + ࠵?࠵? &’ 0 = 8.55࠵? / ࠵? − 9.81࠵?/࠵? # ∗ ࠵? &’ ࠵? &’ = 8.55࠵?/࠵? 9.81࠵?/࠵? # ࠵? &’ = 0.87࠵? ࠵? ",( # = ࠵? ",% # + 2࠵?࠵? " 0 = (8.55࠵? / ࠵?) # −2 ∗ 9.81࠵?/࠵? # ∗ ࠵? ",&’ ࠵? = (8.55࠵? / ࠵?) # 2 ∗ 9.81࠵?/࠵? # ࠵? ",&’ = 3.72࠵? ! ! = ! ! + !" ! = ! ! ! + 0 . 5 ! ! ! ! ! ! = ! ! ! + 2 !" Uniform acceleration equations θ = 40 ° v R = 13.3m/s v y = 8.55m/s v x = 10.19m/s Range d x v y = 0m/s 2.2m a y = -9.8m/s 2 a x = 0m/s 2 y - vertical up v y v x t up t down ∆ d y_up ∆ d y_down Projectile Motion • Calculate total apex height (takeoff + thrown height) • From the apex, calculate t down – Consider direction ࠵? " = ࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵? ℎ࠵?࠵?࠵?ℎ࠵? + ࠵? ",&’ ࠵? "_*+,- = 2.2࠵? + 3.72࠵? ࠵? "_*+,- = 5.92࠵? ࠵? " = ࠵? % ࠵? + 1/2࠵?࠵? # −5.92࠵? = 0 + 1 2 ∗ −9.81࠵?/࠵? # ∗ ࠵? *+,- # ࠵? *+,- = 2 ∗ 5.92࠵? 9.81࠵?/࠵? # ࠵? *+,- = 1.10࠵? ! ! = ! ! + !" ! = ! ! ! + 0 . 5 ! ! ! ! ! ! = ! ! ! + 2 !" Uniform acceleration equations θ = 40 ° v R = 13.3m/s v y = 8.55m/s v x = 10.19m/s Range d x v y = 0m/s 2.2m a y = -9.8m/s 2 a x = 0m/s 2 y - vertical down t up = 0.87s ∆ d y,up = 3.72m t down ∆ d y_down

9 The previous hip, knee, and ankle joint center locations were captured during a jump shot by the basketball player in the figure. Based on the figure and using the previously calculated projection angle, What is the basketball player’s resultant takeoff velocity if the player reached 0.89m above the ground during the shot? During the shot, the basketball player lands on an opposing player’s foot. If the opponent’s shoe and foot are 0.11m tall when on the ground, what is the basketball player’s resultant landing velocity ? What is the player’s total time in the air before landing on the opponent’s foot? How much horizontal distance (range) did the player cover while in the air? 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 ?