2. Newton laws Application Projectile Motion

3. First of All, What is 2-D Motion? Before, we talked about motion in one dimension (just the x axis) Now we are investigating things that can move it 2 dimensions (x and y)! y x 2D Motion x 1D Motion

4. 2-D Motion Part: Projectile Motion A “projectile” is an object on which the only force acting is gravity. –These are things that fly through the air! Usually they are thrown/fired/etc. but are not things like rockets, which still have force pushing them as they fly through the air! As always, force of gravity is only in the down direction.

5. The Dimensions are Independent Since 2D motion is complicated, it is helpful to think of the x and y dimensions separately when we analyze motion. This is because the x and y dimensions are completely independent. This lets us use all of our favorite 1D equations when solving problems!

6. Notice that after this idiot jumps off the cliff his velocity in the x dimension does not change. BUT……His velocity in the y dimension is increasing due to gravity. The same is true for this cannonball:

7. Gravity What force projectiles to accelerate? –Gravity! In which dimension does gravity exist? –The vertical (y)! The acceleration is always 9.81 m/s 2 downward. This due to the sum of forces. Since the force is only in the y dimension this means that the acceleration is always zero for the x dimension.

8. Horizontal Velocity Since we have established that acceleration only exists in the y dimension and the acceleration in the x dimension is zero, what does this mean about the velocity of a projectile in the x dimension? –It is always constant! This needs to be remembered (by YOU)!

9. Horizontal demo

10. Angled Projectile Motion Occasionally, you will encounter problems where the projectile is initially fired at an angle. –This means that the initial velocity will have a horizontal and a vertical component. You will need to calculate these components in order to fill in your givens table on each problem.

11. Vector Components In order to find the components of a vector (like velocity) you will need to use those timeless Trigonometric Functions.

12. Vector Components To solve for v x, we will use cosine because it is the adjacent side and we have the hypotenuse. To solve for v y, use the same process but with sine. Θ=30 o v 0 =50m/s vyvy vxvx

13. As the object accelerates What will happen to the Vertical component, v y as the object is thrown by gravity? –It will change. On the way up it will decrease and on the way down it will increase. Will the v y ever equal 0? –Yes at the highest point. What will happen to the horizontal component, v x as the object is thrown by gravity? –It will remain unchanged. Will the v x ever equal 0? –No unless it started at 0, meaning it was dropped straight down. Θ=30 o v 0 =50m/s vyvy vxvx

14. Solving Problems We are going to take some time to go over the problem solving process for these projectile motion problems. The best part is that you don’t have to learn any complicated equations… we will use the ones you already know from the last unit! It’s easy!

15. Givens Of course, the first step to solving any problem is to list your givens. Always begin by drawing a diagram representing the motion. And remember to include your arrows for initial velocity, acceleration, displacement, and the positive direction. Remember, we are separating the variables into the two dimensions. You will write out the same givens table as before, but this time it will be a different set for each dimension (x or y)! x y v o = v = a x = t = x = v o = v = a y = t = y = + a v d + v d

16. Equations The equations you will use are the same ones. You will just be using the variables from one dimension (x or y) at a time.

17. Time In projectile motion, we will treat the x and y variables as completely different sets. This means the givens we use for one set cannot be used with the other! However, Time is a variable that is always the same for both x and y. x y v o = 25 m/s v = 25 m/s a = 0 m/s 2 t = 4.0 s x = 190 m v o = 0 m/s v = 48 m/s a y = 10.0m/s 2 t = 4.0 s y = 22 m

18. Sample Problem A cannon is placed at the edge of a cliff so the barrel is 60 meters above the ground. The cannon fires a projectile horizontally with a velocity of 95 m/s. How far from the base of the cliff does the projectile land? As always, we will start by drawing a diagram of the situation. Remember to label any givens and include a direction vector for v, d, a, and + y = 60 m x = ?? m vyvy V x =95m/s ayay + + After you have drawn your picture, fill in your Givens table. x y v o = v = a = t = x = v o = v = a y = t = y = 95m/s 0m/s 2 ? ? 0m/s ? 10.0m/s 2 ? 60 m And of course our unknown variable is x. Always circle the unknown in the table.

19. d y = 60 m d x = ?? m vyvy V x =95m/s ayay + + x y v o = v = a = t = x = v o = v = a y = t = y = Sample Problem A cannon is placed at the edge of a cliff so the barrel is 60 meters above the ground. The cannon fires a projectile horizontally with a velocity of 95 m/s. How far from the base of the cliff does the projectile land? ? 95m/s 0m/s 2 ? 0m/s ? 10.0m/s 2 ? 60 m Now determine the equations you will need to solve the problem. Remember, we will keep the sets of variables completely separate (except time)!!! x = v o t + ½ at 2 x = (95)(3.5) + ½ (0)(3.5) 2 x = 332.5 m y = v o t + ½ at 2 t = √(2y / a) t = √[(2)(60) / (10.0)] = 3.5 s This is the only equation we can use to solve for x. But we are missing time! We need to use the variables in the y- dimension to find time first. 3.5 s

20. Thelma and Louise drive their car off the edge of the grand canyon at a speed of 30 m/s and die 3.4 seconds later. (a) how tall is the canyon at that spot? Answer: 57.8 m (b) how far away from the bottom of the canyon do they land? Answer: 102 m Practice Problem

21. Which of the following does not effect the hang time of a projectile? A. Angled fired B. Vertical velocity C. Horizontal velocity D. Height it was fired at.

22. Sample Problem A soccer ball is kicked at an angle of 70 o from the horizontal with a velocity of 20m/s. What is the range of the soccer ball? (how far away does it land) V=20m/s vxvx vyvy Θ=70 o + x y v i = v f = a x = t = d x = v i = v f = a y = t = d y = + a dxdx -10.0 m/s 2 ?? 0 m? 0 m/s 2 As always, start by drawing a diagram, including all vectors. Begin to fill in what you can on your givens table ? ?? ? We need to use trig functions to solve for v ix and v iy v ix = VcosΘ v ix =(20)cos(70) v ix =6.84 m/s v iy = VsinΘ v iy =(20)sin(70) v iy =18.8 m/s +18.8m/s 6.84m/s -18.8m/s And as always circle your unknown variable.

23. Sample Problem A soccer ball is kicked at an angle of 70 o from the horizontal with a velocity of 20.0m/s. What is the range of the soccer ball? (how far away does it land) x y v i = 6.84 m/s v f = 6.84 m/s a x = 0 m/s 2 t = ? d x = ? v i = +18.8 m/s v f = -18.8 m/s a y = -10.0m/s 2 t = ? d y = 0 m V=20m/s vxvx vyvy Θ=70 o + + a dxdx d x = v ix t + ½ at 2 d x = (6.84)(t) + ½ (0)t 2 d x = (6.84)(3.76) + 0 d x = 25.7 m vf = vi + at t = (v f – v i ) / a t = (-18.8 – 18.8) / (-10.0) t = 3.76 s