# Projectile Motion: Horizontally Launched Ball vs Dropped Ball

In this demo, we compare the motion of a ball that is horizontally launched and one that is dropped. Through careful observation, we determine which ball hits the ground first. This is chapter amp on Laserdisk side B, available at only select stores.

## About Projectile Motion: Horizontally Launched Ball vs Dropped Ball

PowerPoint presentation about 'Projectile Motion: Horizontally Launched Ball vs Dropped Ball'. This presentation describes the topic on In this demo, we compare the motion of a ball that is horizontally launched and one that is dropped. Through careful observation, we determine which ball hits the ground first. This is chapter amp on Laserdisk side B, available at only select stores.. The key topics included in this slideshow are . Download this presentation absolutely free.

## Presentation Transcript

1. Demo • Horizontally launched ball vs dropped ball • Which one hits the ground first? • Laserdisk side B chp 11 & 14

2. At only 5’7”, he won the 1986 NBA Slam Dunk Championship where he defeated many high- flying artists that were more than a foot taller than him, including Dominique Wilkins.

3. In Fact: Spud Webb holds the vertical jump world record at an amazing 1.25 m.

4. So just how much “hang time” did he get? ∆y = ½at 2 g t = √2∆y g = √2(-1.25) - 9.8 = .50 s 1 second record hang time double it

5. Okay, that’s great… but Jordan had a running start! Hang time depends only on the jumper’s vertical speed at launch… not horizontal. Horizontal speed remains constant but vertical speed undergoes acceleration.

9. Projectile Motion Parabolic path Range (  x): the horizontal distance a projectile travels Height (  y): the vertical distance a projectile travels.

10. Read directions and answer 1-5 on page 83 • Then, Do 1-3 on page 84 • Finish Lab with collection of data and answer # 2

11. FBD: ball in trajectory Fg a FBD: ball on table Fg a = 0 F N 1.

15.

16. The Basics •  Fnet y = Fg • ma = mg = -9.8 m/s 2 therefore, a = -9.8 m/s 2 • V iy = V fy ; ( velocity is changing ) • Use the 3 kinematic equations   F x = F net x = 0  a = 0 m/s 2  From the kinematic equations, we get  V ix = V fx , constant horizontal velocity  Therefore ,  x = v ix t Vertical Horizontal Remember FBD: Freefall Fg a • time (t y ) • time (t x )

17. The Formulas •  y = ½ at 2 +v iy t • v fy = v iy + a y t • v fy 2 = v iy 2 + 2a y  y  x= ½ at 2 +v ix t • so,  x = v ix t 0 Vertical Horizontal Remember….. • v fx = v ix + a x t • v fx 2 = v ix 2 + 2a x  x 0 0 These last two equations do not help in finding new unknowns.

18. Here’s how to get the answer • First draw a diagram. • Write out your givens. • First solve for the horizontal. You know that a in the horizontal Is 0. And that the change in x=Vit. And that means 9.05=Vit. We need t so let’s see if we can find it in the vertical. • You know that a in the vertical is -9.8 and that the Vi=0. Also change in y is equal to .5at^2+Vit. So plug that in and you get -1=.5(-9.8)t^2 + (0)t So you solve for t and you get t=.452. Then you have t so you can solve for the horizontal vi and you get 19.9m/s. Easy! • Vertically • V iy = ?,  y = ? • a y = -9.8m/s 2 •  y = 1/2 at 2 +v iy t • v f = v i + a y t • v f 2 = v i 2 + 2a y  y • Horizontally • V ix = ?,  x = ? • a x = 0 •  x = v ix t • Picture • Given • Unknowns  y  x

19. Warm-up • In the absence of air resistance, why does the horizontal component of velocity for a projectile remain constant, and why does only the vertical component change? • Now design a lab to determine the initial velocity of a tennis ball shot horizontally from a toy cannon.

20. FLAT TRAJECTORY • A bullet shot from a very high velocity rifle may travel one hundred feet or more without dropping at all. A) true B) false Explain.

21. THE DROPPING BULLET • At the same time that a high-speed bullet is fired horizontally from a rifle, another bullet is simply dropped from the same height. Which bullet strikes the ground first? a) The dropped bullet b) The fired bullet c) Both strike at the same time Explain.

22. Some more fun problems A dodge ball is kicked horizontally at 10 m/s off a 20 m cliff. Find: a) The time it takes for it to hit the ground. b) How far from the cliff it lands. t = 2.02 s; Δ x = 20.2 m

23. Remember…….. • First draw a diagram. • Write out your givens. • First solve for the horizontal. You know that a in the horizontal Is 0. And that the change in x=Vit. And that means 9.05=Vit. We need t so let’s see if we can find it in the vertical. • You know that a in the vertical is -9.8 and that the Vi=0. Also change in y is equal to .5at^2+Vit. So plug that in and you get -1=.5(-9.8)t^2 + (0)t So you solve for t and you get t=.452. Then you have t so you can solve for the horizontal vi and you get 19.9m/s. Easy! • Vertically • V iy = ?,  y = ? • a y = -9.8m/s 2 •  y = 1/2 at 2 +v iy t • v f = v i + a y t • v f 2 = v i 2 + 2a y  y • Horizontally • V ix = ?,  x = ? • a x = 0 •  x = v ix t • Picture • Given • Unknowns  y  x

24. Projectile Practice 1. A marble is fired via sling shot at a barn wall 15 m away. The glass spheroid travels with an initial horizontal velocity of 33 m/s. A. How long does it take to reach the barn? B. How far has it dropped vertically when it strikes the wall? t = .4545 s; Δ y = -1.01 m

25. Check this out: Check this out: How can I find out how fast the object is going when it hits the ground? Or what angle the object strikes the ground? V fx V fy V R

26. 2 . Freddie Frog takes a swan dive from a 13- meter board. He leaves the board at 6.5m/s. If his initial velocity is horizontal, at what angle and at what speed does Freddie hit the water. You Try!!! You Try!!! Answer: v f = 17.23 m/s @ 67.84 ° below the horizontal

28. Launched Horizontally Launched Horizontally Launched at an Angle Launched at an Angle V iy = 0 m/s V iy ≠ 0 m/s

32. More fun problems • A cannon shoots a ball upwards at an angle of 30 degrees from the horizontal. If the V i is 100 m/s. a) Find the horizontal v ix b) the vertical v iy c) the time d) the ∆ x 30 º *Assume ball was launched at ground level t = 10.204 s; Δ x = 883.7 m

33. The Answer • Since the Vi is 100 , and the angle is 30 ° , then the vertical Vi is 100sin30 and the horizontal Vi is 100cos30 . • Solving for horizontal . You know that the Δ x is equal to Vit . We need t to solve this, so let’s come back to it later. • Solving for vertical . We know that the Δ y is equal to 0.5at 2 +Vit . So plug the numbers and we get the Δ y is equal to 0.5(-9.8)t 2 + 100sin30t . We also know the Δ y is 0 because the ball lands where it starts. Factor out a t , and then solve for it and you get t=10.2 seconds . • Now we have t so we can get the horizontal displacement which turns out to be 100cos30*10.2 which is 883.35 meters .

34. A zoo keeper invents a canon to shoot bananas to a monkey who is too shy to come down from the trees and eat. If the monkey does not move, should the zoo keeper aim at, above, or below the monkey? Critical Thinking Critical Thinking

35. If there was no gravity the zoo keeper could aim right at the monkey… but in real life there is gravity so the zoo keeper should aim above the monkey!

36. If the monkey lets go of the branch at the instant the zoo keeper shoots the banana, should the zoo keeper aim at, above, or below the monkey to get the monkey the banana in mid-air?

37. If he aims above the monkey… whoops! He should aim at the monkey… anyone know why?

38. The Answer: The Answer: Because of gravity, the banana (like any projectile) will fall below this direct straight line. How far below? As far as the monkey falls in the same amount of time. So the monkey and the banana will fall the same vertical distance in the same time, and mid- air contact is made.

39. But doesn’t it matter how hard he shoots the banana? Nope, they’ll still meet in mid-air just at different vertical positions high-speed shot medium-speed shot low-speed shot

40. Steph hits a volleyball straight up from a height of 0.80 m with an initial velocity of +7.5 m/s. a) How high will the volleyball go? b) How long will it take for the volleyball to reach its maximum height?

41. Spud was yelling at his friend when suddenly his gum flew horizontally out of his mouth. If his mouth was 1.6 m above the ground and the gum flew a distance of 8 m, a) determine the initial velocity of the gum

42. A football is kicked off a tee at an angle of 40 ° to the horizontal with a velocity of 3 m/s. a) Determine the time the football is in the air b) Determine the range the football travels.

43. 3. An arrow is shot at an angle and hits a target 20 m away and is in the air for 2.5 seconds. a. Find the initial velocity of the arrow. b. Find the maximum height of the arrow. Assume ∆y = 0 V i 14.63 m/s @ 56.8 º above horizontal , Δ y = 7.66 m

44. Projectile Warm up A Football player is trying to kick a field goal from 35 meters out. If he kicks at an angle of 40 degrees, what initial velocity must be given to the ball? The goalpost is 3 meters high. 19.69 m/s