Body weight Ground Reaction Force 70° a. What is the player's vertical take-off velocity, assuming vertical velocity began at 0 m/s? b. Calculate the height reached by the player's center of mass. NEWTON'S LAW OF GRAVITATION F = G m1 x m2 d² where G = gravitational constant = 6.7 x 10-11 Nm²/kg² WEIGHT IS W= M * G GRAVITY IS G=-9.81 M*S^-2 Net Force = mass x acceleration [F = ma ΣF.t=m(v, -v;) ΣF.t=mv-mv; Impulse (J) Force * Time о Total Net Impulse (EJ) = ΣF x time IMPULSE-MOMENTUM IS ΣJ = mAx=m(vt - Vi) NEWTON'S 2ND LAW: LAW OF ACCELERATION Fr= resistive force ER propulsive force ΣF = FR - Fr ΣΕ = ma MOMENTUM IS (p = mv) UNITS ARE: kg⚫m/s VELOCITY TAKE OFF: Vtakeoff = = ΣJ/m COMPUTE JUMP HEIGHT KNOWING V-TAKEOFF WITH EQUATION OF PROJECTILE MOTION: vr² = vi² + 2ad d=-v²/2g FRICTION FORCE: CONSERVATION OF MOMENTUM Σmvbefore = Σmvafter m1*V1 + m2*V2 = (m1 + m2)xt Impulse J = Force (N) * Time (s) AND THE UNITES ARE N⚫s. • Total jump impulse = the total area under the force-time curve (F-dt) Jump (net) impulse = Total impulse - Body Weight impulse IMPULSE/MOVEMENTUM RELATIONSHIP coefficient of *Force Normal normal reaction force (1 to plane of contact) F Static Friction Maximum static friction force static friction COEFFICIENT OF FRICTION: F₁ = u N PRESSURE: pressure= force area ΣΕ = ma ΣF •t=m(v₁-v;) ΣF=m V₁ - Vi t CF tmv, -1 -mvi • Heel contact Barefoot walking • F-1 BW (e.g., 800 N) ⚫ critical contact area can vary depending on surface: • The left side of this equation is the: •Impulse (Ns) Force (N) * Time (s) •The right side of this equation is: •Momentum (kgm/s) = Mass (kg) * Velocity (m/s) SI Units: N/m² (Pascal) English units: lb/in² (psi) pressure-force area walking on a smooth surface est. area: 25 cm² = 0.0025 m² - stepping on a small rock (diam=.25cm=.1in) est. area: 0.05 cm² = 0.000005 m² P = 320,000 Pa P = 320 Kilo Pa (KPa) P = 160,000,000 Pa P = 160,000 KPa stepping on a tack (diam .05cm.02in) est. area: 0.0020 cm² = 0.0000002 m² P = 400,000,000 Pa P = 400.000 KPa

Body weight Ground Reaction Force 70° a. What is the player's vertical take-off velocity, assuming vertical velocity began at 0 m/s? b. Calculate the height reached by the player's center of mass. NEWTON'S LAW OF GRAVITATION F = G m1 x m2 d² where G = gravitational constant = 6.7 x 10-11 Nm²/kg² WEIGHT IS W= M * G GRAVITY IS G=-9.81 MS^-2 Net Force = mass x acceleration [F = ma ΣF.t=m(v, -v;) ΣF.t=mv-mv; Impulse (J) Force Time о Total Net Impulse (EJ) = ΣF x time IMPULSE-MOMENTUM IS ΣJ = mAx=m(vt - Vi) NEWTON'S 2ND LAW: LAW OF ACCELERATION Fr= resistive force ER propulsive force ΣF = FR - Fr ΣΕ = ma MOMENTUM IS (p = mv) UNITS ARE: kg⚫m/s VELOCITY TAKE OFF: Vtakeoff = = ΣJ/m COMPUTE JUMP HEIGHT KNOWING V-TAKEOFF WITH EQUATION OF PROJECTILE MOTION: vr² = vi² + 2ad d=-v²/2g FRICTION FORCE: CONSERVATION OF MOMENTUM Σmvbefore = Σmvafter m1V1 + m2V2 = (m1 + m2)xt Impulse J = Force (N) * Time (s) AND THE UNITES ARE N⚫s. • Total jump impulse = the total area under the force-time curve (F-dt) Jump (net) impulse = Total impulse - Body Weight impulse IMPULSE/MOVEMENTUM RELATIONSHIP coefficient of Force Normal normal reaction force (1 to plane of contact) F Static Friction Maximum static friction force static friction COEFFICIENT OF FRICTION: F₁ = u N PRESSURE: pressure= force area ΣΕ = ma ΣF •t=m(v₁-v;) ΣF=m V₁ - Vi t CF tmv, -1 -mvi • Heel contact Barefoot walking • F-1 BW (e.g., 800 N) ⚫ critical contact area can vary depending on surface: • The left side of this equation is the: •Impulse (Ns) Force (N) Time (s) •The right side of this equation is: •Momentum (kgm/s) = Mass (kg) * Velocity (m/s) SI Units: N/m² (Pascal) English units: lb/in² (psi) pressure-force area walking on a smooth surface est. area: 25 cm² = 0.0025 m² - stepping on a small rock (diam=.25cm=.1in) est. area: 0.05 cm² = 0.000005 m² P = 320,000 Pa P = 320 Kilo Pa (KPa) P = 160,000,000 Pa P = 160,000 KPa stepping on a tack (diam .05cm.02in) est. area: 0.0020 cm² = 0.0000002 m² P = 400,000,000 Pa P = 400.000 KPa