![The Horizontal Motion:
At any time t, the projectile’s horizontal displacement 𝑥 − 𝑥0 from an initial
position 𝑥0 is given by
𝑥 − 𝑥0 = 𝑣0𝑥 𝑡 +
1
2
𝑎𝑥𝑡2
Where 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑎𝑙𝑜𝑛𝑔 𝑥 − 𝑎𝑥𝑖𝑠, 𝑎𝑥 = 0
Using 𝑣0𝑥 = 𝑣0 cos 𝜃0 we can write
𝑥 − 𝑥0 = (𝑣0cos 𝜃0) 𝑡 ……….. (1)
At any time t, the projectile’s horizontal velocity 𝑣0𝑥 = 𝑣𝑥
The Vertical Motion:
At any time t, the projectile’s vertical displacement y − 𝑦0 from an initial
position 𝑦0 is given by
𝑦 − 𝑦0 = 𝑣0𝑦 𝑡 −
1
2
𝑔𝑡2 [ where, 𝑎𝑦 = −𝑔]
𝑦 − 𝑦0 = (𝑣0sin 𝜃0) 𝑡 −
1
2
𝑔𝑡2
[ where, 𝑣0𝑦 = 𝑣0 sin 𝜃0]
……………………… (2)](https://image.slidesharecdn.com/physics1lesson2midspring24-25-250412203829-0362f57a/85/Physics-1_LESSON-2-Mid_Spring-24-25-pdf-8-320.jpg)
![❑ Equations for the horizontal range and the maximum
horizontal range of a projectile:
The horizontal range R of the projectile is the horizontal distance the
projectile has traveled when it returns to its initial height (the height at which
it is launched). That is 𝑥 − 𝑥0 = 𝑅 𝑤ℎ𝑒𝑛 𝑦 − 𝑦0 = 0.
Using 𝑥 − 𝑥0 = 𝑅 𝑖𝑛 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 1 𝑎𝑛𝑑 𝑦 − 𝑦0 = 0 𝑖𝑛 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 (2), we get
𝑅 = (𝑣0cos 𝜃0) 𝑡
And 0 = (𝑣0sin 𝜃0) 𝑡 −
1
2
𝑔𝑡2
[From equation (1)]
[From equation (2)]
𝑜𝑟 (𝑣0 sin 𝜃0) 𝑡 =
1
2
𝑔𝑡2
𝑜𝑟 𝑡 =
2𝑣0 sin 𝜃0
𝑔
Therefore, 𝑅 = (𝑣0cos 𝜃0)
2𝑣0 sin 𝜃0
𝑔
=
𝑣0
2
(2 sin 𝜃0 cos 𝜃0)
𝑔
𝑅 =
𝑣0
2
sin 2𝜃0
𝑔
Caution: This equation does not give the
horizontal distance traveled by a projectile
when the final height is not the launch height.
……(3)](https://image.slidesharecdn.com/physics1lesson2midspring24-25-250412203829-0362f57a/85/Physics-1_LESSON-2-Mid_Spring-24-25-pdf-11-320.jpg)
![The value of R is maximum in equation (3) when sin 2𝜃0 = 1
𝑜𝑟 2𝜃0 = sin−1 1
𝑜𝑟 2𝜃0 = 900
[𝑠𝑖𝑛𝑐𝑒 sin−1
1 = 900
]
𝜃0 = 450
The Effects of the Air (in the projectile motion):
The launch angle is 60° and the launch speed is 44.7 m/s.](https://image.slidesharecdn.com/physics1lesson2midspring24-25-250412203829-0362f57a/85/Physics-1_LESSON-2-Mid_Spring-24-25-pdf-12-320.jpg)
![1. [Chap 4 - problem 21]: A dart is thrown horizontally with an initial speed of 10 m/s
toward point P, the bull’s-eye on a dart board. It hits at point Q on the rim,
vertically below P, 0.19 s later. (a)What is the distance PQ? (b) How far away from
the dart board is the dart?
2. A projectile is fired horizontally from a gun that is 45.0 m above flat ground,
emerging from the gun with a speed of 250 m/s. (a) How long does the projectile
remain in the air? (b) At what horizontal distance from the firing point does it
strike the ground? (c) What is the magnitude of the vertical component of its
velocity as it strikes the ground?
3. A soccer ball is kicked from the ground with an initial speed of 19.5 m/s at an
upward angle of 45°. A player 55 m away in the direction of the kick starts running
to meet the ball at that instant. What must be his average speed if he is to meet the
ball just before it hits the ground?
4. In Fig. 4-34, a stone is projected at a cliff of height h
with an initial speed of 42.0 m/s directed at angle
𝜽𝟎 = 60.0° above the horizontal. The stone strikes
at A, 5.50 s after launching. Find (a) the height h
of the cliff, (b) the speed of the stone just before
impact at A, and (c) the maximum height H reached above the ground
Let’s Practice !!](https://image.slidesharecdn.com/physics1lesson2midspring24-25-250412203829-0362f57a/85/Physics-1_LESSON-2-Mid_Spring-24-25-pdf-16-320.jpg)
Physics 1_LESSON 2 (Mid_Spring 24-25).pdf
- 1. LESSON 2 BOOK CHAPTER 4 Projectile Motion
- 2. Look at the pictures !!!
- 3. Projectile Motion: A particle moves in a vertical plane with some initial velocity Ԧ 𝑣0 but its acceleration is always the freefall acceleration Ԧ 𝑔, which is downward. Such a particle is called a projectile (meaning that it is projected or launched), and its motion is called projectile motion. Figure: The trajectory of an idealized projectile. Examples: A batted baseball, a thrown football, a package dropped from an airplane, and a bullet shot from a rifle are all projectiles.
- 4. Sketch of the path taken in projectile motion: Figure: An object launched into the air at the origin of a coordinate system and with launch velocity Ԧ 𝑣0 at angle 𝜃0. The motion is a combination of vertical motion (constant acceleration) and horizontal motion (constant velocity), as shown by the velocity components.
- 5. Animation of projectile motion
- 6. At a certain instant, a fly ball has velocity 𝑣 = 5 Ƹ 𝑖 − 4.9 Ƹ 𝑗 (the x axis is horizontal, the y axis is upward, and 𝑣 is in meters per second). Has the ball passed its highest point? Check your understanding
- 7. The adjacent figure is a stroboscopic photograph of two golf balls. One ball is released from rest and the other ball is shot horizontally at the same instant. The golf balls have the same vertical motion, both falling through the same vertical distance in the same interval of time. The fact that one ball is moving horizontally while it is falling has no effect on its vertical motion; that is, the horizontal and vertical motions are independent of each other.
- 8. The Horizontal Motion: At any time t, the projectile’s horizontal displacement 𝑥 − 𝑥0 from an initial position 𝑥0 is given by 𝑥 − 𝑥0 = 𝑣0𝑥 𝑡 + 1 2 𝑎𝑥𝑡2 Where 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑎𝑙𝑜𝑛𝑔 𝑥 − 𝑎𝑥𝑖𝑠, 𝑎𝑥 = 0 Using 𝑣0𝑥 = 𝑣0 cos 𝜃0 we can write 𝑥 − 𝑥0 = (𝑣0cos 𝜃0) 𝑡 ……….. (1) At any time t, the projectile’s horizontal velocity 𝑣0𝑥 = 𝑣𝑥 The Vertical Motion: At any time t, the projectile’s vertical displacement y − 𝑦0 from an initial position 𝑦0 is given by 𝑦 − 𝑦0 = 𝑣0𝑦 𝑡 − 1 2 𝑔𝑡2 [ where, 𝑎𝑦 = −𝑔] 𝑦 − 𝑦0 = (𝑣0sin 𝜃0) 𝑡 − 1 2 𝑔𝑡2 [ where, 𝑣0𝑦 = 𝑣0 sin 𝜃0] ……………………… (2)
- 9. At any time t, the projectile’s vertical velocity 𝑣𝑦 = 𝑣0 sin 𝜃0 − 𝑔𝑡 And we can express 𝑣𝑦 2 𝑎𝑠 𝑣𝑦 2 = 𝑣0 sin 𝜃0 2 − 2 𝑔 𝑦 − 𝑦0 ❑ Show that the path of a projectile is a parabola. From equation (1) we can write 𝑡 = 𝑥 − 𝑥0 𝑣0 cos 𝜃0 Using the value of t in equation (2), we get 𝑦 − 𝑦0 = 𝑣0 sin 𝜃0 𝑥−𝑥0 𝑣0 cos 𝜃0 − 1 2 𝑔 𝑥−𝑥0 𝑣0 cos 𝜃0 2
- 10. For simplicity, we let 𝑥0 = 0 𝑎𝑛𝑑 𝑦0 = 0. 𝑦 = tan 𝜃0 𝑥 − 1 2 𝑔 𝑥 𝑣0 cos 𝜃0 2 Therefore, the equation becomes ………………… (3) Where 𝜃0, 𝑔 𝑎𝑛𝑑 𝑣0 𝑎𝑟𝑒 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑠. Equation (3) is of the form 𝑦 = 𝑎𝑥 ∓ 𝑏𝑥2 , 𝑤ℎ𝑒𝑟𝑒 𝑎 𝑎𝑛𝑑 𝑏 𝑎𝑟𝑒 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑠. This is the equation of a parabola, so the path is parabolic.
- 11. ❑ Equations for the horizontal range and the maximum horizontal range of a projectile: The horizontal range R of the projectile is the horizontal distance the projectile has traveled when it returns to its initial height (the height at which it is launched). That is 𝑥 − 𝑥0 = 𝑅 𝑤ℎ𝑒𝑛 𝑦 − 𝑦0 = 0. Using 𝑥 − 𝑥0 = 𝑅 𝑖𝑛 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 1 𝑎𝑛𝑑 𝑦 − 𝑦0 = 0 𝑖𝑛 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 (2), we get 𝑅 = (𝑣0cos 𝜃0) 𝑡 And 0 = (𝑣0sin 𝜃0) 𝑡 − 1 2 𝑔𝑡2 [From equation (1)] [From equation (2)] 𝑜𝑟 (𝑣0 sin 𝜃0) 𝑡 = 1 2 𝑔𝑡2 𝑜𝑟 𝑡 = 2𝑣0 sin 𝜃0 𝑔 Therefore, 𝑅 = (𝑣0cos 𝜃0) 2𝑣0 sin 𝜃0 𝑔 = 𝑣0 2 (2 sin 𝜃0 cos 𝜃0) 𝑔 𝑅 = 𝑣0 2 sin 2𝜃0 𝑔 Caution: This equation does not give the horizontal distance traveled by a projectile when the final height is not the launch height. ……(3)
- 12. The value of R is maximum in equation (3) when sin 2𝜃0 = 1 𝑜𝑟 2𝜃0 = sin−1 1 𝑜𝑟 2𝜃0 = 900 [𝑠𝑖𝑛𝑐𝑒 sin−1 1 = 900 ] 𝜃0 = 450 The Effects of the Air (in the projectile motion): The launch angle is 60° and the launch speed is 44.7 m/s.
- 13. Problem 22 (Book chapter 4): A small ball rolls horizontally off the edge of a tabletop that is 1.20 m high. It strikes the floor at a point 1.52 m horizontally from the table edge. (a) How long is the ball in the air? (b) What is its speed at the instant it leaves the table? 𝑦 − 𝑦0 = −1.2 𝑚 𝜃0 = 00 𝑎𝑛𝑑 𝑣0 =? 𝑥0 𝑥 𝑥 − 𝑥0 = 1.52 𝑚 𝑡 =? Answer: 𝑦 − 𝑦0 = (𝑣0sin 𝜃0) 𝑡 − 1 2 𝑔𝑡2 −1.20 = 0 − 4.9𝑡2 (a) We know −1.20 = (𝑣0 𝑠𝑖𝑛 00) 𝑡 − 4.9𝑡2 𝑡 = 1.2 4.9 = 0.495 𝑠 (b) We know 𝑥 − 𝑥0 = (𝑣0cos 𝜃0) 𝑡 1.52 = (𝑣0cos 00)( 0.495) 1.52 = (𝑣0cos 00)( 0.495) 1.52 = (𝑣0)(1)(0.495) 𝑣0 = 1.52 0.495 = 3.07 𝑚/𝑠
- 14. You throw a ball toward a wall at speed 25.0 m/s and at angle 40.0° above the horizontal (as shown in the figure). The wall is distance d= 22.0 m from the release point of the ball. (a) How far above the release point does the ball hit the wall? What are the (b) horizontal and (c) vertical components of its velocity as it hits the wall? (d) When it hits, has it passed the highest point on its trajectory? Problem 32 (Book chapter 4): Answer:(a) We know Given 𝑣0 = 25 𝑚/𝑠 ; 𝜃0 = 400 𝑥 − 𝑥0 = 𝑑 = 22 𝑚 𝑎 𝑦 − 𝑦0 =? 𝑏 𝑣𝑥 =? 𝑎𝑛𝑑 𝑐 𝑣𝑦 =? 𝑑 𝐷𝑖𝑑 𝑡ℎ𝑒 𝑏𝑎𝑙𝑙 𝑝𝑎𝑠𝑠 𝑡ℎ𝑒 ℎ𝑖𝑔ℎ𝑒𝑠𝑡 𝑝𝑜𝑖𝑛𝑡? 𝑦 − 𝑦0 = (𝑣0sin 𝜃0) 𝑡 − 1 2 𝑔𝑡2 𝑦 − 𝑦0 = (25)(sin 400 ) 𝑡 − 4.9 𝑡2 𝑦 − 𝑦0 = 25 0.6428 𝑡 − 4.9 𝑡2 𝑦 − 𝑦0 = 16.07𝑡 − 4.9 𝑡2 To find t we use the following formula, 𝑥 − 𝑥0 = (𝑣0cos 𝜃0) 𝑡 𝑡 = 𝑥 − 𝑥0 𝑣0 cos 𝜃0 = 22 25 𝑐𝑜𝑠400 = 22 (25)(0.7660) 𝑦 − 𝑦0 = 16.07 1.149 − 4.9 1.149 2 = 18.46 − 6.469 = 11.99 𝑚 Therefore, 𝑡 = 1.149 𝑠
- 15. (b) We know 𝑣𝑥 = 𝑣0𝑥 = 𝑣0 cos 𝜃0 = 25 cos 400 = 25 0.766 = 19.15 𝑚/𝑠 (c) We know 𝑣𝑦 = 𝑣0 sin 𝜃0 − 𝑔𝑡 = 25 sin 400 − (9.8)(1.149) 𝑣𝑦 = 25 0.6428 − 11.26 = 4.81 m/s (d) Since 𝑣𝑦 𝑖𝑠 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒, 𝑡ℎ𝑎𝑡 𝑖𝑠, 𝑣𝑦 > 0, 𝑡ℎ𝑒 𝑏𝑎𝑙𝑙 𝑑𝑖𝑑 𝑛𝑜𝑡 𝑟𝑒𝑎𝑐ℎ 𝑡𝑜 𝑡ℎ𝑒 ℎ𝑖𝑔ℎ𝑒𝑠𝑡 𝑝𝑜𝑖𝑛𝑡 𝑜𝑛 ℎ𝑖𝑡𝑡𝑖𝑛𝑔 𝑡ℎ𝑒 𝑤𝑎𝑙𝑙.
- 16. 1. [Chap 4 - problem 21]: A dart is thrown horizontally with an initial speed of 10 m/s toward point P, the bull’s-eye on a dart board. It hits at point Q on the rim, vertically below P, 0.19 s later. (a)What is the distance PQ? (b) How far away from the dart board is the dart? 2. A projectile is fired horizontally from a gun that is 45.0 m above flat ground, emerging from the gun with a speed of 250 m/s. (a) How long does the projectile remain in the air? (b) At what horizontal distance from the firing point does it strike the ground? (c) What is the magnitude of the vertical component of its velocity as it strikes the ground? 3. A soccer ball is kicked from the ground with an initial speed of 19.5 m/s at an upward angle of 45°. A player 55 m away in the direction of the kick starts running to meet the ball at that instant. What must be his average speed if he is to meet the ball just before it hits the ground? 4. In Fig. 4-34, a stone is projected at a cliff of height h with an initial speed of 42.0 m/s directed at angle 𝜽𝟎 = 60.0° above the horizontal. The stone strikes at A, 5.50 s after launching. Find (a) the height h of the cliff, (b) the speed of the stone just before impact at A, and (c) the maximum height H reached above the ground Let’s Practice !!
- 17. Thank You