Presentation on theme: "Falling Objects & Terminal Velocity"— Presentation transcript:
1 Falling Objects & Terminal Velocity Notes p.24Although the acceleration due to gravity is 9.8 ms-2, downwards, objects do not maintain this acceleration in air because ________ ________________ acts upwards.airresistanceIMPORTANT EXPLANATION of TERMINAL VELOCITYAs 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 ________ ________ .airresistanceincreasesdecreaseairresistanceweightconstantvelocityterminalvelocity
2 ExampleSketch 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 forcesTerminal Velocity.Parachute opensVelocity 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
3 Conservation of Energy Notes p.25From Nat 5 you should recall the following key energy equations:E p = mghEk = ½ mv2EW = FdE = P tEp =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.
4 Worked ExampleA 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.1stEk = ?Ek = ½ mv2m = 0.02 kg= 0.5 x 0.02 x 62v = 6 ms-1= 0.36 J2nd Kinetic energy changes to work done against friction!d = EW / FEw = 0.36 J= / 30F = 30 N= md = ?= 1.2 cm