Motion in 2 Dimensions Projectile Motion A projectile

Motion in 2 -Dimensions

Projectile Motion A projectile is given an initial force and is then (assuming no air resistance) is only acted on by gravity. The path it takes is called the trajectory. Path is a parabolic shape.

Horizontal Projectiles Horizontal - cause is due to the throw or movement. vh is constant. The force only acts for an instant. No force = no acceleration. dh = vht

Vertical - force acting on the object is only gravity. vv changes because gravity is acting on the object. dv = vit + 1/2 gt 2 Time is interchangeable between horizontal and vertical directions

Motion Map

Practice A ball is rolled horizontally off of a flat roof. It hits the ground 2. 3 m away from the wall. If the ball was in the air for 1. 2 seconds, what was the horizontal velocity of the ball just before it rolled off of the roof? How tall was the roof?

Projectiles Launched at an Angle o When launched at an angle the projectile has an initial vertical component as well as a horizontal component. o 1 st step should be to find the components of the velocity vector. o These act independently from each other o Solving for maximum height use half of the trajectory o vertical velocity = 0 and t = 1/2 total t o Range - how far the object travels horizontally. o Flight Time (hang time) - the total time the object is in the air.

Motion Map Vertical Velocity Horizontal Velocity Vertical Acceleration and Fg

Maximum Range • A projectile will have the maximum range when it is fired at an angle of 450 • Two angles that are the same value away from 45 will have the same range. • EX 30 and 60, 10 and 80 • Animation 1

Practice • A ball is launched with an initial velocity of 4. 47 m/s at an angle of 66 o above the horizontal. • a) What is the maximum height the ball attained? vvi=visin vvi=4. 47 m/s sin(66) = 4. 08 m/s vh=vicos vh=4. 47 m/s cos(66) = 1. 81 m/s

We need to know how long the ball takes to reach its maximum height. vvf=vvi + at 0 = 4. 08 m/s + (-9. 8 m/s 2)t t =. 416 s Now we know how long it takes to reach maximum height. dv = vvit + 1/2 gt 2 dv = (4. 08 m/s)(. 416 s) + 1/2(-9. 8 m/s 2)(. 416 s)2 dv =. 849 m

£How long does it take to return to the height it was launched from? (total time) Time to max height = 0. 416 s Total time = 2 x 0. 416 s Total time = 0. 832 s

What is the range? • • Range(R) = dh dh = vht R = (1. 81 m/s) (. 832 s) R = 1. 51 m

#2 £A soccer player kicks a ball into the air at an angle of 36 degrees. The initial velocity is 30 m/s. ¤How long is the ball in the air? ¤What is the horizontal distance traveled by the ball? ¤What is the maximum height reached by the ball?

#3 £An arrow is shot at 30 degrees with a velocity of 49 m/s and hits a target. ¤What is the maximum height attained by the arrow? ¤The target is at the height the arrow was shot from. How far away is the target?