Mechanics Chapter 6 Motion due to Gravity 𝒏

6.1 Objects falling from height You already know from Chapter 3 objects falling under the influence of gravity has an acceleration of 𝑔=10 𝑚𝑠 −2 . This is excluding air resistance, for now. Since the acceleration is constant, you can use equations from section 1.5 to answer questions with gravity then.

6.1 Objects falling from height Example 1 At a swimming pool a girl steps from a diving board 4 m above the surface of the water. How fast is she moving when her feet hit the water?

6.1 Objects falling from height Example 2 A brick is dislodged from the top of a tall apartment building. A resident on a 10th floor balcony sees it passing, and a second later hears it hit the ground. Each story has a height of 2.5 m. How tall is the building and how fast is it moving when it hits the ground?

6.2 Objects projected upwards Gravity is still acting on an object while in the air. But if the object is moving upwards, the force is pulling it down, so gravity is now a deceleration, still with value g g = -10

6.2 Objects projected upwards Example 1 A ball is thrown vertically upwards and rises a height of 12.8 m. Find the speed with which it was thrown, and its velocity when it has risen 11 meters.

6.3 Motion on a sloping plane We are now going to look again at sloping planes, like in chapter 4 and 5. This time the object will be moving either upwards or downwards on the slope. Also, there will be friction, both smooth and rough introduced.

6.3 Motion on a sloping plane Example 1 A path runs up a hillside, at an angle of α to the horizontal, such that sin 𝛼 =0.6 𝑎𝑛𝑑 cos 𝛼=0.8 . A block is struck and starts to move up the path at a speed of 12 𝑚𝑠 −1 . The path is icy so friction can be ignored. Find how far up the path the block moves, the speed with which it hits the curb on its return, and the time it is in motion.

6.3 Motion on a sloping plane Example 2 Same example as #1 but now the ice has melted and there is a coefficient of friction between the block and the path of 0.45

6.4 Vertical Motion with Air resistance So far whenever an object is falling we haven’t considered air resistance. Air resistance in the air is like friction on a surface. There is a point where the air resistance balances the weight and the acceleration has maxed out. This is called the terminal speed.

6.4 Vertical Motion with Air resistance For air resistance, it is equal to the velocity squared time some constant k 𝑎𝑖𝑟 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒=𝑘 𝑣 2

6.4 Vertical Motion with Air resistance Three people step out of an aircraft, and fall vertically before opening their parachutes. The first has mass 80 kg falls upright with a terminal speed of 50 𝑚𝑠 −1 . The second has mass 120 kg and falls upright also. The third has mass 70 kg, and falls horizontally, which gives her air resistance by a factor of 12. Find the terminal velocity of the second and third people to jump out.

6.4 Vertical Motion with Air resistance A cannonball is projected vertically upwards from a mortar with an initial velocity of 40𝑚𝑠 −1 . The mortar is situated at the edge of a cliff 100 meters above the sea. On the way down, the cannonball just misses the cliff. In vertical fall the cannonball would have a terminal speed of 50𝑚𝑠 −1 . Calculate the acceleration of the cannonball just after it leaves the mortar barrel and at the highest point of its path. Draw graphs to compare the actual motion with the motion predicted if there were no air resistance.