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Projectile Motion: Finding Max Height

Projectile Motion: Finding Max Height. E. D. F. C. G. B. H. A. I. X up. X down. X up = X down. Properties of Projectile Motion. Projectile motion is symmetrical around the “maximum height” of the projectile's path. The Acceleration is always 9.8m/s 2 Downward!!!!!.

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Projectile Motion: Finding Max Height

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  1. Projectile Motion: Finding Max Height

  2. E D F C G B H A I X up X down Xup = Xdown Properties of Projectile Motion • Projectile motion is symmetrical around the “maximum height” of the projectile's path The Acceleration is always 9.8m/s2 Downward!!!!! SpeedA = SpeedI SpeedB = SpeedH SpeedC = SpeedG SpeedD = SpeedF From A to E speed slows down Since Vx is constant and X up = X down Then Time up = Time down From E to I speed speeds up Speed is slowest at E

  3. Changes in a Projectile’s Velocity As a projectile reaches maximum height the velocity slows down and flattens out. Vy goes to 0 m/s at a rate of 9.8m/s2, Vx is a constant At max height Vy is ALWAYS 0 m/s!! The acceleration is, however, 9.8 m/s2 downward 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

  4. Finding Maximum Height when there is symmetry • 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.

  5. Old Example…New Application 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)

  6. From before we know tup = 1.02 sec (Half of range, Hmax) (17.665,Hmax) Vi = 20 m/s @ 30O (35.33 m, 0m) H max ttotal = 2.04 sec (0m, 0m) 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

  7. Can we find the time to reach maximum height if the path is not symmetrical? • 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.

  8. Using Velocity to find time up. • 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

  9. Lets revisit an old friend 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

  10. Maximum height in 1 step • 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.

  11. Time independent equation • 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

  12. One more time 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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