 # Falling Objects & Terminal Velocity

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Falling Objects & Terminal Velocity
Notes p.24 Although the acceleration due to gravity is 9.8 ms-2, downwards, objects do not maintain this acceleration in air because ________ ________________ acts upwards. air resistance IMPORTANT EXPLANATION of TERMINAL VELOCITY As a falling object speeds up, the _____ _________ acting upwards ________ . This causes the unbalanced force and acceleration to ________ . Eventually the object will reach a speed where the upwards force due to ___ ________ balances the _______ of the object. From this point, the object will maintain a ________ ______ . We call this ________ ________ . air resistance increases decrease air resistance weight constant velocity terminal velocity

Example Sketch a velocity – time graph to illustrate the motion of a skydiver from the moment she jumps from the plane. Your graph should clearly indicate the point at which she opens her parachute. Velocity (ms-1) Time (s) ‘a’ = 0 … balanced forces Terminal Velocity. Parachute opens Velocity decreases as air resistance bigger than weight. But air resistance decreases with speed, so rate of deceleration decreases. ‘a’ decreases as air resistance increases with speed. New “terminal velocity” when forces balance again. initially ‘a’ = 9.8 ms-2

Conservation of Energy
Notes p.25 From Nat 5 you should recall the following key energy equations: E p = mgh Ek = ½ mv2 EW = Fd E = P t Ep = m = g = h = Ek = v = potential energy (in J) Ew = F = d = E = P = t = work done (in J) mass (in kg) force (in N) gravitational field strength (in Nkg-1) distance (in m) energy transferred (in J) power (in W) height (in m) kinetic energy (in J) time (in s) velocity (in ms-1) Revision questions for Higher Physics, page 26, Q. 1 – 14.

Worked Example A 20 g dart is travelling at 6 ms-1 when it strikes a dart board. The dartboard exerts an average frictional force of 30 N on the dart. Determine the dartboard thickness required to bring the dart to rest. 1st Ek = ? Ek = ½ mv2 m = 0.02 kg = 0.5 x 0.02 x 62 v = 6 ms-1 = 0.36 J 2nd Kinetic energy changes to work done against friction! d = EW / F Ew = 0.36 J = / 30 F = 30 N = m d = ? = 1.2 cm

Revision questions for Higher Physics, page 26, Q. 1 – 14.