2.  Projectile motion is symmetrical around the “maximum height” of the projectile's path Since V x is constant and X up = X down Then Time up = Time down A B C D E F G H I Speed A = Speed I Speed B = Speed H Speed C = Speed G Speed D = Speed F From A to E speed slows down From E to I speed speeds up Speed is slowest at E X up X down X up = X down The Acceleration is always 9.8m/s 2 Downward!!!!!

3. As a projectile reaches maximum height the velocity slows down and flattens out. V y goes to 0 m/s at a rate of 9.8m/s 2, V x is a constant At max height V y is ALWAYS 0 m/s!! The acceleration is, however, 9.8 m/s 2 downward As the projectile falls from maximum height the velocity increases and becomes more vertical V y increases at a rate of 9.8m/s 2 V X is constant

4.  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? Set up: 1 st Get a Visual: Draw the situation Range (0m, 0m) (range, 0m) 2 nd 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) V i = 20 m/s @ 30 O H max (Half of range, Hmax)

6. Range (0m, 0m) (35.33 m, 0m) V i = 20 m/s @ 30 O H max (Half of range, Hmax) t total = 2.04 sec (17.665,H max ) t up = 1.02 sec Pos Yf = Pos Yi + V Yi t +(½)a y t 2 H max = 0 m + (10 m/sec) (1.02 sec) +(1/2)(-9.8 m/sec 2 ) (1.02 sec) 2 H max 5.1 m

7.  If the path is not symmetrical we can not simply divide the total time by two because t up = t down  In stead we need to look at the velocity of the projectile to find the time.  We know that at maximum height V y must be 0 m/s.

8.  Since we are looking for time and we do not know the final position we can use the velocity equation to find time  V f = V i + at  But looking at the Y axis only  V fy = V iy + a g t  0 m/sec = V i *Sin  + a g t up  So we can say that : t up = -(Vi*Sin  /a g  And now that we know t up we can use the position equation for the Y axis and find H max

9. 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? Range (0m, 0m) (range, 0m) V i = 20 m/s @ 30 O H max (Half of range, Hmax) V fy = V iy + a g t 0 m/sec= 20m/sec*Sin   + (-9.8 m/sec 2 )t up t up = 1.02 sec t up = -(20m/sec*Sin    / (-9.8 m/sec 2 ) Pos Yf = Pos Yi + V Yi t +(½)a y t 2 H max = 0 m + (10 m/sec) (1.02 sec) +(1/2)(-9.8 m/sec 2 ) (1.02 sec) 2 H max 5.1 m

10.  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.  We start with the time independent formula ◦ V f 2 = V i 2 + 2a(Pos f – Pos i )  We now apply it to the Y axis only  V yf 2 = V yi 2 + 2a y (Pos yf – Pos yi )  For H max we know ◦ V yf = 0 m/s ◦ V yi = V i Sin  ◦ a =a g ◦ Pos f = H max ◦ Pos i = starting height {H i } (This is usually 0 m)  So we can know say  (0 m/s)2 = (V i Sin  )2 + 2(a g )(H max – H i )  This means that ◦ H max = [-(V i Sin  ) 2 /(2a g )] + Hi

12. 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? Range (0m, 0m) (range, 0m) V i = 20 m/s @ 30 O H max (Half of range, Hmax) V yf 2 = V yi 2 + 2a y (Pos yf – Pos yi ) (0 m/s) 2 = (20m/s*Sin 30) 2 + 2(-9.8 m/s 2 )(H max – 0m) H max = -(20m/s*Sin 30) 2 / [2*(-9.8 m/s 2 )] H max 5.1 m