Projectiles IB Revision

Gravity does not act sideways gravity makes it accelerate downwards The ball moves with a constant horizontal velocity The ball accelerates downwards at 10 ms -2

Horizontal motion does not alter the effect of gravity which acts vertically A ball dropped on a moving train will take the same time to reach the floor as one dropped on a stationary train A ball travelling sideways and one that is dropped vertically will fall at the same rate Ouch! Projectiles

Question A ball is kicked horizontally at 10 ms -1 from the top of a cliff that is 45 m high. How long does it take to reach the ground? Does this question deal with horizontal or vertical motion? S = ut +1/2 at 2 what is the initial vertical velocity. U = 0 t = 3s How far from the base of the cliff does the ball land? 3 x 10 = 30m what is its velocity (magnitude and direction) when it hits the ground vertical v = u + at Pythagorus says 31.6 ms -1 10ms -1 45m

30 10 Pythagorus says 31.6 ms -1 Resolving Components In this case you added the horizontal and vertical vectors to find the resultant You often want to do the opposite V = 20 ms o Horizontal component = v cos 30 = 17.3 ms -1 Vertical component = v sin 30 = 10 ms -1

Projectile launched at an angle The path of the projectile is a parabola It has constant horizontal velocity equal to v cos  Initial vertical velocity is v sin  and there is an acceleration of 9.8ms -2 Bang! V V cos  V sin 

Max Height If the projectile is fired at 15 ms -1 at an angle of 60 o to the ground calculate the maximum height Initial vertical velocity = vsin  = 13 ms -1 Vertical velocity at max height = 0 v 2 = u 2 + 2as s = 13 2 /19.6 = 8.6 m Vertical velocity at max height = 0

Time of Flight If the ground is horizontal then time to go up = time to go down Fired at 15ms -1 at 60 o to the ground v = u + at t = 2 x 13/9.8 = 2.7 s V = 15ms -1  = 60 o

Range Range = horizontal speed x time 15cos60 x 2.7 = 20.2 m Range = 20.2 m It can be proved that maximum range occurs when launched at 45 o V = 15ms -1  = 60 o

KE and PE KE at launch = 1/2mv 2 = 1/2 x 2 x 15 2 = 225 J PE at launch = 0 KE at max height = 1/2 mv 2 = 1/2 x 2 x = 56 J PE = mgh = 2 x 9.8 x 8.6 = 169 J Total E = = 225J Use energy to find v at a height of 5m 1/2 mv 2 = mgh v = 11.3 ms -1 Mass = 2kg v = 15 ms -1 at 60 o

Projectile velocity Find v at a height of 5m without using energy vertical v 2 = u 2 + 2as v = x 9.8 x 5 = 8.4 ms -1 horizontal v = 7.5ms -1 Pythagorus gives v = 11.3 ms -1 v = 15 ms -1 at 60 o