Presentation on theme: "Regents Physics – Mr. Rockensies"— Presentation transcript:
1Regents Physics – Mr. Rockensies Projectile MotionRegents Physics – Mr. Rockensies
2Projectiles 3 Types: Vertical, Horizontal, Inclined An object is launched with an initial velocity in the vertical direction.At the peak of the object’s flight, v = 0 m/s.Once the object returns to the same height it was released from, its velocity will be the same as its initial.v = 0 m/sVertical
3viObject starts out with an initial velocity in the horizontal direction.Acceleration due to gravity gives the object a velocity in the vertical direction.Horizontal
4An object is launched with a velocity that has vertical and horizontal components. These components can be resolved by setting up a right triangle.viInclined
5Independence of Motion The vertical and horizontal motion (free fall) does not affect the horizontal motion (constant velocity)
6The Horizontal Projectile viIn order to find the vertical distance travelled, we have to separate the horizontal motion from the vertical motion.dy
7Vertical Motion (y)viy = 0 m/say = -9.8 m/s2dy = heightto find the time, use:dy = viyt + ½ayt2dy = ½ayt2Horizontal Motion (x)vix = viax = 0 m/s2dx = vixt + ½axt2dx = vixtNote: - ax = 0!!!! Gravity affects only vertical motion.- time of flight is the same regardless of vi, a dropped bullet and a fired bullet hit the ground at the same time.
8Practice Problem 1A pool ball leaves a 0.60-meter high table with an initial horizontal velocity of m/s. Predict the time required for the pool ball to fall to the ground and the horizontal distance between the table's edge and the ball's landing location.
9Practice Problem 2A soccer ball is kicked horizontally off a 22.0-meter high hill and lands a distance of 35.0 meters from the edge of the hill. Determine the initial horizontal velocity of the soccer ball.
10Trajectory – The path of a Projectile For Horizontal projectile motion: -ax = 0 m/s2! Gravity affects only vertical motion -time of flight is the same regardless of vi, a dropped bullet and a fired bullet hit the ground at the same time. Horizontal Motion Mythbusters Bullet Drop
11The Inclined Projectile Initial velocity must be resolved into components (viy and vix)Vy = 0 m/s at topviyviθvix
12t obtained from vertical to find range: dx = vixt + ½ axt2 How do we solve this??Horizontal Motion (x)Vix = vicosθax = 0 m/s2t obtained from vertical to find range:dx = vixt + ½ axt2Vertical Motion (y)viy = visinθ very important!!ay = m/s2vfy = 0 m/s at topto find time:use vfy = viy + ayt to get time to top, then double it.to find max height:vfy2 = viy2 + 2aydy
13Practice Problem #1An airplane traveling 1001 m above the ocean at 125 km/h is going to drop a box of supplies to shipwrecked victims below.a. How many seconds before the plane is directly overhead should the box be dropped?b. What is the horizontal distance between the plane and the victims when the box is dropped?
14Practice Answer #1Vertical viy = 0 m/s ay = -9.8 m/s2 dy = 1001 m t = ? d = ½ at m = ½ (9.8 m/s2)t2 t = 14.3 sHorizontalvix = 125 km/hdx = ?t = ?ax = 0 m/s2dx = vixtdx = (125 km/h)(14.3 s)dx = ( km/s)(14.3 s)dx = km = m
15Practice Problem #2Herman the human cannonball is launched from level ground at an angle of 30° above the horizontal with an initial velocity of 26 m/s. How far does Herman travel horizontally before reuniting with the ground?
16Horizontal vix = 26(cos30) ax = 0 m/s2 dx = ? t = ? dx = vixt Practice answer #2Vertical viy = 26(sin30) ay = -9.8 m/s2 vf = 0 m/s t = ? vf = viy + at 0 = 26(sin30) + (-9.8 m/s2)t -13 = -9.8 m/s2t t = 1.33 sHorizontalvix = 26(cos30)ax = 0 m/s2dx = ?t = ?dx = vixtdx = 26(cos30)(2.66 s)dx = 60.0 m
17Effect of Launch Angle Complementary angles produce the same range. Maximum range is at 45 degrees launch angle.Higher angle leads to longer “hang time”