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2-Dimensional Kinematics. 2-Dimensional Kinematics. 2-Dimensional Kinematics. Projectile Motion Projectiles Launched Horizontally A projectile is an object shot through the air.

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## 2-Dimensional Kinematics

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**2-Dimensional Kinematics**Projectile Motion Projectiles Launched Horizontally A projectile is an object shot through the air. A projectile’s path through space is called its trajectory. The shape of the trajectory is essentially a parabola.**2-Dimensional Kinematics**Practice Problem: A movie director is shooting a scene that involves dropping a stunt dummy out of an airplane and into a swimming pool. The plane is 10.0 m above the ground, traveling at a velocity of 22.5 m/s in the positive x direction. The director wants to know where in the plane’s path the dummy should be dropped so that it will land in the pool. What do you tell him? Also, find the instantaneous velocity of the stunt dummy as it hits the water.**2-Dimensional Kinematics**Practice Problem #2: A pelican flying along a horizontal path drops a fish from a height of 5.4 m while traveling 5.0 m/s. How far does the fish travel horizontally before it hits the water below? Also, give both the horizontal and vertical components of the velocity of the fish before it enters the water.**2-Dimensional Kinematics**Practice Problem #3: A cat chases a mouse across a 1.0 m high table. The mouse steps out of the way, and the cat slides off the table at a speed of 5.0 m/s. Where does the cat strike the floor?**2-Dimensional Kinematics**Projectiles Launched at an Angle**2-Dimensional Kinematics**v0 = the initial velocity of the projectile (note: This is NOT initial velocity in just the x- or y-direction. It is the overall initial velocity in the direction of the projectile’s launch angle.) = the launch angle of the projectile g = acceleration due to gravity (9.80 m/s2)**2-Dimensional Kinematics**R = where: R = the range of the projectile (how far it traveled)**2-Dimensional Kinematics**t = where: t = the time aloft (hang time) of the projectile**2-Dimensional Kinematics**ymax = where: ymax = the maximum height reached by the projectile**2-Dimensional Kinematics**Problem #1: The narrowest strait on earth is Seil Sound in Scotland, which lies between the mainland and the island of Seil. The strait is only about 6.0 m wide. Suppose an athlete wanting to jump “over the sea” leaps at an angle of 35° with respect to the horizontal. What is the minimum initial speed that would allow the athlete to clear the gap? Neglect air resistance.**2-Dimensional Kinematics**Problem #2: A flying fish leaps out of the water with a speed of 15.3 m/s. Normally these fish use wing-like fins to glide about 40 m before reentering the ocean, but in this case the fish fails to use its “wings” and so only travels horizontally about 17.5 m.At what angle with respect to the water’s surface does the fish leave the water?**2-Dimensional Kinematics**Problem #3: A ball is launched at 4.5 m/s at 66° above the horizontal. What are the maximum height and flight time of the ball?**2-Dimensional Kinematics**Problem #4: A zookeeper finds an escaped monkey hanging from a light pole. Aiming her tranquilizer gun at the monkey, the zookeeper kneels 10.0 m from the light pole, which is 5.00 m high. The tip of her gun is 1.00 m above the ground. The monkey tries to trick the zookeeper by dropping a banana, then continues to hold onto the light pole. At the moment the monkey releases the banana, the zookeeper shoots. If the tranquilizer dart travels at 50.0 m/s, will the dart hit the monkey or the banana?**2-Dimensional Kinematics**Problem #5: In a scene in an action movie, a stuntman jumps from the top of one building to the top of another building 4.0 m away. After a running start, he leaps at an angle of 15 with respect to the flat roof while traveling at a speed of 5.0 m/s. If the other building is 2.5 m shorter than the building from which he jumps, will he make it to the other roof?**2-Dimensional Kinematics**Problem #6: A golfer can hit a golf ball a horizontal distance of over 300 m on a good drive. What maximum height would a 301.5 m drive reach if it were launched at an angle of 25.0 to the ground? (Hint: At the top of its flight, the ball’s vertical velocity will be zero.)**2-Dimensional Kinematics**Problem #7: Salmon often jump waterfalls to reach their breeding grounds. Starting 2.00 m from a waterfall 0.55 m in height, at what minimum speed must a salmon jumping at an angle of 32.0 leave the water to continue upstream?**2-Dimensional Kinematics**Problem #8: A quarterback throws the football to a receiver who is 31.5 m down the field. If the football is thrown at an initial angle of 40.0 to the ground, at what initial speed must the quarterback throw the ball? What is the ball’s highest point during its flight?**2-Dimensional Kinematics**Frame of Reference & Relative Motion If you are moving at 80 km/h north and a car passes you going 90 km/h, to you the faster car seems to be moving north at 10 km/h. Someone standing on the side of the road would measure the velocity of the faster car as 90 km/h toward the north. Velocity measurements depend of the frame of reference of the observer.**2-Dimensional Kinematics**Suppose you are in a school bus that is traveling at a velocity of 8 m/s in a positive direction. You walk with a velocity of 3 m/s toward the front of the bus. If a friend of yours is standing on the side of the road watching as you go by, how fast would your friend say that you are moving?**2-Dimensional Kinematics**Considering only the forward (horizontal) direction, how fast would an observer on the ground say the balloon was travelling? Would the pilot have a different answer?**2-Dimensional Kinematics**Relative Velocity: va/b + vb/c = va/c This formula holds true even when the relative motions are in two dimensions (and not in one dimension, as in the bus example).**2-Dimensional Kinematics**A pilot wants to fly a plane east at a velocity of 400.0 km/hr with respect to the ground. A 50.0 km/hr wind is blowing southward. With respect to the air, what velocity must the pilot maintain?**2-Dimensional Kinematics**Ana and Sandra are riding on a ferry boat that is traveling east at a speed of 4.0 m/s. Sandra rolls a marble with a velocity of 0.75 m/s north, straight across the deck of the boat to Ana. What is the velocity of the marble relative to the water?**2-Dimensional Kinematics**A polar bear swims 2.60 m/s south relative to the water. The bear is swimming against a current that moves 0.78 m/s at an angle of 40.0° north of west, relative to Earth. How long will it take the polar bear to reach the shore, which is 5.50 km to the south?**2-Dimensional Kinematics**The world’s fastest current is in Slingsby Channel, Canada, where the speed of the water reaches 30.0 km/h. Suppose a motorboat crosses the channel perpendicular to the bank at a speed of 18.0 km/h relative to the bank. Find the velocity of the motorboat relative to the water.**2-Dimensional Kinematics**Circular Motion Consider an object moving in a circle at a constant speed, like a ball being whirled on the end of a string, or that ride at the fair where you sit in a swing and they spin you around really fast. Once you hit a constant velocity, are you accelerating in that swing any more? Since the speed doesn’t change, you might say no – acceleration is constant. Remember, however, that acceleration is defined as change in velocity, not just change in speed. Since you are constantly changing direction as you spin in that swing, you must also be constantly accelerating…**2-Dimensional Kinematics**Uniform circular motion is the movement of an object at a constant speed around a circle with a fixed radius.**2-Dimensional Kinematics**You find an object’s acceleration by subtracting the velocity vectors.**2-Dimensional Kinematics**The average acceleration is in the same direction as v, toward the center of the circle. No matter what the time interval, the length of the acceleration vector remains the same. The direction of the acceleration vector (for an object in uniform circular motion) always points toward the center of the circle. The acceleration of such an object is therefore called center-seeking or centripetal acceleration, ac.**2-Dimensional Kinematics**Without worrying too much about how it is derived, the formula to calculate centripetal acceleration is: ac =**2-Dimensional Kinematics**One way to measure the speed of an object moving in a circle is to measure its period, T, which is the time needed for the object to make one complete revolution. During one revolution, the object travels a distance equal to the circle’s circumference, 2r. So the object’s velocity could be defined as: v = =**2-Dimensional Kinematics**If you substitute this expression for velocity in the equation for centripetal acceleration, you get:

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