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Projectile : an object acted upon only by the force of gravity Putting our SUVAT equations to good use

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Learning Objectives : 1.To understand the concept of projectiles 2.To be able to complete projectile calculations successfully NOTE you can ignore “projectiles” altogether and just consider it normal SUVAT work Book Reference : Pages 126-127 A fun example.... [SoccerBall]

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What sort of problems do we need to solve? Problems with only a vertical aspect “Releasing an object from a hot air balloon” Problems with both horizontal and vertical aspects “Putting a golf ball off of the top of a cliff” 101 possible questions about distance, speed, time etc etc Both horizontally and vertically

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Starting points.... 1.We need to consider vertical &horizontal components of motion & they are independent 2.The acceleration involved is always g (downwards) & only affects the vertical component 3.Any horizontal velocity is constant and unaffected by g 4.We’ll consider up as positive and down as negative Watch the following telly if you are not convinced! [02_Parabolic_motion_and_uniformly_accelerated_linear_motion]

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Note : Horizontally there is no change in velocity, (1 square per photo all the way across) Note : Vertically, the distance travelled between each photo increase (acceleration) Note : Vertically, the motion of the dropped ball and the thrown ball is identical

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Vertical Projection Only 1.As the name suggests, the object only travels vertically 2.Vertical velocity at any moment is given by a variation on our 1 st SUVAT equation v = u - gt 3.Our displacement at any moment is given by a variation on our 3 rd SUVAT equation y = ut -½gt 2 4.If there is an initial positive vertical projection (upwards). then the vertical velocity will be zero at the highest point. 5.Remember to use other SUVAT equations as required

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Horizontal Projection 1.It starts with a perfectly horizontal motion 2.As it accelerates vertically downwards its path becomes steeper and steeper 3.Speed of horizontal projection affects how far away it lands 4.But does not affect how long it stays in the air We can watch the telly again if you are not convinced!

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Horizontal Projection If the object is projected with a horizontal velocity u then after t seconds.... Horizontal component of its displacement x = ut Vertical component of its displacement y = ½gt 2 There is no initial vertical component

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u ½gt 2 ut Displacements at time t

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At time t there are both horizontal and vertical velocities: v x = U v y = -gt And the combined speed/velocity will be given by Pythagoras: = √(v x 2 + v y 2 )

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vyvy vxvx √(v x v y ) √(v x 2 + v y 2 ) We can use trigonometry to find θ the direction of our combined velocity For example tan θ = v y / v x θ

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A worked example.... An object is projected at a horizontal velocity of 15 ms -1 from the top of a 35m tower. Calculate the following a)How long it takes to reach the ground (2.67s) b)How far it travels horizontally (40m) c)It’s speed at impact (30.2 ms -1 )

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