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Projectile motion is symmetrical around the “maximum height” of the projectile's path The Acceleration is always 9.8m/s2 SpeedA = SpeedI Downward!!!!! E SpeedB = SpeedH D SpeedC = SpeedG F SpeedD = SpeedF C G B H A I X up X down Xup = Xdown Since Vx is constant and X up = X From A to E speed slows down down From E to I speed speeds up Then Time up = Time down Speed is slowest at E As a projectile reaches maximum height the velocity At max height Vy is ALWAYS 0 m/s!! slows down and flattens The acceleration is, however, out. 9.8 m/s2 downward Vy goes to 0 m/s at a rate of 9.8m/s2, Vx is a constant As the projectile falls from maximum height the velocity increases and becomes more vertical Vy increases at a rate of 9.8m/s2 VX is constant If a projectile’s path is symmetrical then The Time to reach maximum height equals one half of the total time of the projectiles motion. Knowing the time to reach maximum height we can use the position equation (the same equation we always used for projectile motion) to find the maximum height of the projectile. A soccer ball is kicked from ground level with a speed of 20 m/s and an angle of 30 degrees. What is the ball’s Maximum Height? (Half of range, Hmax) Vi = 20 m/s @ 30O (range, 0m) H max (0m, 0m) Range Set up: 1st Get a Visual: Draw the situation 2nd Break the problem into components: Draw X & Y axis's and label the initial and final points of the projectile. (You can have the origin anywhere) tup = 1.02 sec (Half of range, Hmax) (17.665,Hmax) Vi = 20 m/s @ 30O (35.33 m, 0m) H max (0m, 0m) ttotal = 2.04 sec Range PosYf = PosYi + VYit +(½)ayt2 Hmax = 0 m + (10 m/sec) (1.02 sec) +(1/2)(-9.8 m/sec2) (1.02 sec) 2 Hmax 5.1 m If the path is not symmetrical we can not simply divide the total time by two because tup = tdown In stead we need to look at the velocity of the projectile to find the time. We know that at maximum height Vy must be 0 m/s. Since we are looking for time and we do not know the final position we can use the velocity equation to find time Vf = Vi + at But looking at the Y axis only Vfy = Viy + agt 0 m/sec = Vi*Sinq + agtup So we can say that : tup = -(Vi*Sinq)/ag And now that we know tup we can use the position equation for the Y axis and find Hmax A soccer ball is kicked from ground level with a speed of 20 m/s and an angle of 30 degrees. What is the ball’s Maximum Height? (Half of range, Hmax) Vi = 20 m/s @ 30O (range, 0m) H max (0m, 0m) Range Vfy = Viy + agt PosYf = PosYi + VYit +(½)ayt2 0 m/sec= 20m/sec*Sin30O + (-9.8 m/sec2)tup Hmax = 0 m + (10 m/sec) (1.02 sec) +(1/2)(-9.8 m/sec2) (1.02 sec) 2 tup = -(20m/sec*Sin30O )/ (-9.8 m/sec2) Hmax 5.1 m tup = 1.02 sec Using both the velocity equation to find time up, and the position equation to find maximum height will ALWAYS work. However, it is a two step process. And nice would it be if there was a shorter way of doing this? Well, there is. Any time you have to use both the velocity and position equation together you can use the time independent equation instead. We start with the time independent formula ◦ Vf2 = Vi2 + 2a(Posf – Posi) We now apply it to the Y axis only Vyf2 = Vyi2 + 2ay(Posyf – Posyi) For Hmax we know ◦ Vyf= 0 m/s ◦ Vyi = ViSinq ◦ a =ag ◦ Posf= Hmax ◦ Posi = starting height {Hi} (This is usually 0 m) So we can know say (0 m/s)2 = (ViSinq)2 + 2(ag)(Hmax – Hi) This means that ◦ Hmax = [-(ViSinq)2/(2ag)] + Hi A soccer ball is kicked from ground level with a speed of 20 m/s and an angle of 30 degrees. What is the ball’s Maximum Height? (Half of range, Hmax) Vi = 20 m/s @ 30O (range, 0m) H max (0m, 0m) Range Vyf2 = Vyi2 + 2ay(Posyf – Posyi) (0 m/s)2 = (20m/s*Sin 30)2 + 2(-9.8 m/s2)(Hmax – 0m) Hmax= -(20m/s*Sin 30)2/ [2*(-9.8 m/s2)] Hmax 5.1 m

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posted: | 12/28/2010 |

language: | English |

pages: | 12 |

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Slide Katy ISD Range

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