Motion in Two Dimensions

(Ignore any effects from air resistance) A pickup is moving with a constant velocity and a hunter is sitting in the back pointing his rifle straight up. If he fires the rifle, where will the bullet land? In front of the pickup? Behind the pickup? Inside the pickup?

Two people stand on the edge of a cliff. One throws a rock straight up with a speed of 20 m/s. The other throws an identical rock straight down with a speed of 20 m/s. Which rock will be moving faster when it hits the ground?

A rifle, at a height H above the ground, fires a bullet parallel to the ground. At the same instant and from the same height H, a second bullet is dropped from rest. Which bullet strikes the ground first?

The Earth is rotating about its axis at a speed of 463 m/s at the equator. Considering that speed, when a person throws a ball straight up into the air, why doesn’t he rotate out from under it before the ball falls back down?

While riding in a car traveling down the road, you reach out and drop a cat. What would its pathway to the ground look like to you? What would the cat’s pathway look like to a person standing alongside the road?

Projectiles Projectile: an object that moves through air or space, acted upon only by gravity Range: the horizontal distance traveled by a projectile Trajectory: the pathway of a projectile Gravity provides acceleration in the vertical direction. But….. Assuming no air resistance, there is NOTHING to affect its horizontal velocity so… The horizontal acceleration is ZERO

Drop or Launch horizontally? Which one hits the ground first? Gravity acts the same on both!same

An object dropped from a given height will hit the ground AT THE SAME TIME As an object launched horizontally from the same height. Why? Gravity pulls both of them to the ground at the same rate. Acceleration due to gravity = -9.8 m/s/s

Projectiles Launched Horizontally V = 20 m/s d = v o t + ½ at 2 x (horizontal ) y (vertical) v o = 20 m/s v o = 0 a = 0 a = -10 m/s 2 d x = v o t d y = ½ at 2 H = 5t 2 H = 50m Range = d x Range = 20 x 3.16 = 63.2 m -H = ½ (-10)t 2

Projectiles Launched Horizontally V = 30 m/s d = v o t + ½ at 2 x y V o = 30 m/s Vo = 0 a = 0 a = -10 m/s 2 d x = v o t d y = ½ at 2 H = 5t 2 H = 50m Range = d x Range = 30 x 3.16 = 94.8m

Projectiles Launched Horizontally V = 10 m/s d = v o t + ½ at 2 x y V o = 10 m/s Vo = 0 a = 0 a = -10 m/s 2 d x = v o t d y = ½ at 2 H = 5t 2 H = 50m Range = d x Range = 10 x 3.16 = 31.6m

Projectiles Launched Horizontally d = v o t + ½ at 2 x y V o = 0 m/s Vo = 0 a = 0 a = -10 m/s 2 d x = v o t d y = ½ at 2 H = 5t 2 H = 50m

Projectiles Launched Horizontally V = ? x (horizontal) y (vertical) v o = ? v o = 0 a = 0 a = -10 m/s 2 d x = v o t d y = ½ at 2 H = 5t 2 H = ? Range = d x = 120 m t = 3 s H = 5 x 3 2 = 45 m Vo = d / t = 120 / 3 = 40 m/s

Projectiles Launched Horizontally A cannon ball is shot horizontally at 45 m/s off the top of a cliff 100 m high. How long did it take for the ball to hit the ground? How far did it land from the base of the cliff H = 5t 2 d = v o t H / 5 = t 2 t = 4.47 s d = v o t d = 45 (4.47) = 201 m

A ball rolling at 8 m/s rolls off the top of a 20 m high building. How long will it take to hit the ground? How far will it land from the base of the building? H = 5t 2 d = v o t H / 5 = t 2 t = 2 s d = v o t d = 8 (2) = 16 m

A projectile hits the ground 3 seconds after it was launched horizontally from the top of a hill. If the range of the projectile was 60 m, how fast was the projectile launched? How tall was the hill? H = 5t 2 d = v o t H = 5 (3) 2 = 45 m d = v o t v o = d / t = 60 / 3 = 20 m/s

If a projectile hit the ground 4 seconds after it was launched and the range was 50 m, how fast was the launch velocity? How tall was the hill? H = 5t 2 d = v o t H = 5 (4) 2 = 80 m d = v o t v o = d / t = 50 / 4 = 12.5 m/s

A bullet is shot horizontally from the top of a hill. It strikes the ground in 1.5 seconds with a range of 700 m. How tall was the hill? How fast was the original velocity of the bullet? H = 5t 2 d = v o t H = 5 (1.5) 2 = m d = v o t v o = d / t = 700 / 1.5 = 466 m/s

Projectiles Launched at an Angle The original velocity, v o of the projectile is not entirely horizontal nor entirely vertical, but has COMPONENTS that are both horizontal and vertical. vxvx vyvy vovo

The vertical component of the velocity, v y is affected by gravity and will change at a rate of –10 m/s every second. g = -10m/s 2 The horizontal component of the velocity, v x is NOT affected by gravity and if there is no air resistance, it remains constant. vxvx vyvy vovo

Suppose the vertical component of the velocity, v y, was 50 m/s. What will it be 1 second after the launch? Remember, g = -10 m/s per second V y = 40 m/s V y = 50 m/s

Suppose the vertical component of the velocity, v y, was 50 m/s. What will it be 2 second after the launch? V y = 30 m/s

Suppose the vertical component of the velocity, v y, was 50 m/s. What will it be 3 second after the launch? V y = 20 m/s

Suppose the vertical component of the velocity, v y, was 50 m/s. What will it be 4 second after the launch? V y = 10 m/s

Suppose the vertical component of the velocity, v y, was 50 m/s. What will it be 5 second after the launch? V y = 0 m/s

Suppose the vertical component of the velocity, v y, was 50 m/s. What will it be 6 second after the launch? vxvx V y = -10 m/s vovo

Suppose the vertical component of the velocity, v y, was 50 m/s. How could one find the time it takes until a projectile reaches its highest point? vxvx V y = 50 m/s vovo

Suppose the vertical component of the velocity, v y, was 60 m/s. What will it be 1 second after the launch? 2 s? 3s? 4s? 5s? 6s? 7s? How could one find the time it takes until a projectile reaches its highest point? vxvx V y = 60 m/s vovo

Suppose the vertical component of the velocity, v y, was 55 m/s. What will it be 1 second after the launch? How could one find the time it takes until a projectile reaches its highest point? vxvx V y = 55 m/s vovo

Suppose the vertical component of the velocity, v y, was 40 m/s and the horizontal component of the velocity, v x, was 30 m/s. How long will it be until it reaches its highest point? 4s How much longer until it hits the ground? 4s How far in the horizontal x direction has the projectile traveled during those 8 seconds? In other words, what is its RANGE? Range = d x = v x t = (30 m/s) (8 s) Range = 240 m V x = 30 m/s V y = 40 m/s vovo 240 m

Suppose the vertical component of the velocity, v y, was 60 m/s and the horizontal component of the velocity, v x, was 15 m/s. How long will it be until it reaches its highest point? 6s How much longer until it hits the ground? 6s How far in the horizontal x direction has the projectile traveled during those 12 seconds? In other words, what is its RANGE? d x = v x t = (15 m/s) (12 s) Range = 180 m V x = 15 m/s V y = 60 m/s vovo 180 m

Suppose the vertical component of the velocity, v y, was 30 m/s and the horizontal component of the velocity, v x, was 50 m/s. How long will it be until it reaches its highest point? 3s How much longer until it hits the ground? 3s How far in the horizontal x direction has the projectile traveled during those 6 seconds? In other words, what is its RANGE? d x = v x t = (50 m/s) (6 s) Range = 300 m V x = 50 m/s V y = 30 m/s vovo 300 m

Suppose the vertical component of the velocity, v y, was 45 m/s and the horizontal component of the velocity, v x, was 30 m/s. How long will it be until it reaches its highest point? 4.5s How much longer until it hits the ground? 4.5s How far in the horizontal x direction has the projectile traveled during those 9 seconds? In other words, what is its RANGE? d x = v x t = (30 m/s) (9 s) Range = 270 m V x = 30 m/s V y = 45 m/s vovo 270 m

Suppose the vertical component of the velocity, v y, was 45 m/s and the horizontal component of the velocity, v x, was 30 m/s. How fast was v o ? It’s not 45 m/s. It’s not 30 m/s. How would you find the original velocity, v o ? Use the Pythagorean Theorem! V x = 30 m/s V y = 45 m/s vovo

At what part of the trajectory does a projectile have minimum speed?

At what angle will a projectile go the farthest? 45 degrees

A ball is launched at an angle of 30 degrees and hits a target. At what other angle could the ball be launched and still hit the target? 45 – 30 = 15 degrees = 60 degrees

If you throw a ball STRAIGHT up, what is its velocity at its highest point? ZERO What is its acceleration at its highest point? - 10 m/s 2

If you throw a ball at an angle, what is its velocity at its highest point? Its velocity at its highest point is NOT zero, but is the same as its original x component of its velocity. What is its acceleration at its highest point? -10 m/s 2

Circular Motion

Rotating Turning about an internal axis Revolving Turning about an external axis

Linear speed How far you go in a certain amount of time Miles per hour, meters per second Rotational speed How many times you go around in a certain amount of time Revolutions per minute, rotations per hour

Which horse has a larger linear speed on a merry go round, one on the outside or one on the inside? Outside. Which horse has a greater rotational speed? Neither, all the horses complete the circle in the same amount of time.

How much faster will a horse at TWICE the distance from the center of the circle be moving TWICE the distance means TWICE the speed

The number of revolutions per second is called the frequency, f. Frequency is measured in Hertz, Hz. The time it takes to go all the way around once is called the period, T. Frequency is related to period by f = 1 / T and T = 1 / f

Example A boy twirls a toy airplane around and around at the end of a string. If it takes 2 seconds for the airplane to complete one loop, what is the frequency? f = 1 / T f = 1 / 2 f = 0.5 Hz

How do you find the velocity of an object moving in a circle if it is not directly provided? We know that Velocity = distance / time In circular motion, the distance traveled is all around the circle… the circumference. The circumference = 2  r The time it takes the object to go all the way around the circumference once is called the period, T. So… v = 2  r / T

Example A race car takes 1.5 minutes to go around one lap of a circular track. If the track has a radius of 400 m, how fast was the car traveling? v = 2  r / T v = 2  (400) / (1.5 x 60) v = 27.9 m/s

Uniform Circular Motion, UCM: moving in a circle with a constant speed. Question: Is there a constant velocity when an object moves in a circle with a constant speed? No, the direction changes, therefore the velocity changes. If the velocity changed, the object is actually ACCELERATING even while moving at the same speed.