Projectile Motion

1. Projectile Motion An object given an initial velocity and then left to move freely under gravity is a projectile.

2. A golfer hit ball A. It took 5 seconds to land in the position shown. • He then hit ball B with less force. How long did it take to land in its position? • 4 seconds • 5 seconds • 6 seconds A B

3. The Adventures of Speedy!

4. Horizontal and vertical motion are independent !

5. Horizontal and vertical motion are independent (90°) • There are the two components of the projectile's velocity – horizontal vX and vertical vY . • The horizontal component of it’s velocity does not change. vX is constant Gravity is the only force

7. Projectiles Projectile motion is free fall with an initial horizontal velocity. y x

8. Projectile motion is free fall with an initial horizontal velocity. It combines constant horizontal velocity and accelerated vertical velocity. Horizontal – constant velocity Use equations like s = v x t y Vertical – a simple free fall Use equations like s= ½ gt2 and g = v/t • The horizontal and vertical motions are independent of each other, but they have a common time. • Air resistance is ignored x

9. Caught by slow motion camera: The yellow ball is given an initial horizontal velocity and the red ball is dropped. Both balls fall at the same rate.

10. 3ms-1 A ball is thrown horizontally from a cliff with an initial velocity of 3ms-1. It landed 9 meters away. a. How long did it take to land? b. How high is the cliff? c. What was its velocity just before hitting the ground? The horizontal motion: s = v x t t = 9/3 = 3 s The vertical motion s= ½ gt2 s = ½ x-9.8 x 32 = -44.1m The vertical motion g = v/t v = -9.8 x 3 = -29.4 ms-1 9 m

11. Package drop • The package follows a parabolic path and remains directly below the plane at all times • The vertical velocity changes (faster, faster) • The horizontal velocity is constant!

12. Summary Q • An object is released, 50m above the ground, from a hot air balloon, which is descending vertically at a speed of 4.0ms-1. Calculate: • The velocity of the object at the ground • The duration of descent of the object • The height of the balloon above the ground when the object hits the ground. a. v2 = u2 + 2as v2 = (-4)2 + 2 x-9.8 x -50 v = -31.6ms-1 b. a =(v-u)/t -9.8 = (-31.6--4) / t t = 2.8 s u = -4ms-1 c. After 2.8 seconds, distance travelled by the balloon is 4x2.8 = 11.2m That is 38.8m above the ground 50m v = ?

13. 2. An object is projected horizontally at a speed of 16ms-1 into the sea from a cliff top of height 45.0m. Calculate: • How long it takes to reach the sea • How far it travels horizontally • Its impact vertical velocity. 16ms-1 45.0m • Taking vertical motion • s= ½ gt2 • -45 = ½ x -9.8 x t2 • t = 3 s b. Taking horizontal motion s = v x t s = 16 x 3 = 48m c. The vertical motion g = v/t v = gt v = -9.8 x 3 = -29.4 ms-1

14. 3. A dart is thrown horizontally along a line passing through the centre of a dartboard, 2.3 m away from the point at which the dart was released. The dart hits the dartboard at a point 0.19m below the centre. Calculate: • The time of flight of the dart • Its horizontal speed of projection. • Taking vertical motion • s= ½ gt2 • -0.19 = ½ x -9.8 x t2 • t = 0.197 s = 0.2 s 2.3 m • Taking horizontal motion • s = v x t v = s/t • v = 2.3/0.197 = 11.7 ms-1

15. 4. A parcel is released from an aircraft travelling horizontally at a speed of 120ms-1 above level ground. • The parcel hits the ground 8.5 seconds later. Calculate: • The height of the aircraft above the ground • The horizontal distance travelled in this time by i) the parcel, ii) the aircraft • The speed of impact of the parcel on the ground. • Taking vertical motion s = ½ gt2 • s = ½ x -9.8 x 8.52 = -354 m • b. Taking horizontal motion s = v x t • s = 120 x 8.5 = 1020m • This is the distance travelled by the aircraft and the parcel • c. Taking vertical motion g = v/t • vertical speed v = -9.8 x 8.5 = -83.3 ms-1 • Horizontal speed = 120 ms-1 • (resultant speed)2 = 83.32 + 1202 Resultant speed = 146 ms-1

16. A native hunter in the Amazon is hunting monkeys with a bow and a tranquiliser arrow. The hunter aims at a monkey sitting on a branch. As the hunter releases the arrow the monkey notices the movement and drops off the branch. What happens next? A. The arrow misses, passing above the monkey. B. The arrow hits the monkey. C. You need to know the distance to know what happens. The monkey was a bit too smart for its own good. If the monkey had not dropped from the branch the arrow would have followed a parabolic path, passing under the branch missing the monkey. The arrow's y-position changes in the same way as the monkey's y-position changes when it drops off the branch. Thus the arrow's path curves to meet the monkey.

17. A zookeeper wants to throw a banana to the monkey who hangs from the limb of a tree. This particular monkey has a habit of dropping from the tree the moment that he sees the zookeeper throwing the banana. In the presence of gravity, banana moves in a parabolic path. The monkey also accelerates downward once he lets go of the limb. Both banana and monkey experience the same acceleration. So the monkey will catch the banana.