![Chalkboard Example:
Use the diagram from the previous page
Let v1 = 100.0 m/s [300
up from horizontal]
Determine: a.) time of fight b.) range c.)a.) time of fight b.) range c.)
max height of projectile from ground d.)max height of projectile from ground d.)
velocity of projectile just before it hits thevelocity of projectile just before it hits the
groundground
ANS a.) 10.2 s b.) 884 m c.) 128 ma.) 10.2 s b.) 884 m c.) 128 m
d.) Discussd.) Discuss
It is important to know that the greatestIt is important to know that the greatest
range is obtained whenrange is obtained when θ = 450
andand
greatest time of flight whengreatest time of flight when θ = 900
..](https://image.slidesharecdn.com/u1c-m-l4-pm-in-vert-comp-academic-r-161022010943/85/Grade-11-U1C-L4-Vertical-PM-Academic-4-320.jpg)
![Chalkboard Example
An arrow is shot with an initial velocity ofAn arrow is shot with an initial velocity of 10.0 m/s
[300
Up from Horiz.] from a bridge which isfrom a bridge which is 10.0 m
above the ground. Lettingabove the ground. Letting g = 10.0 m/s2
,,
determine the following:determine the following:
1.1. Time of flightTime of flight
2.2. Maximum height above the groundMaximum height above the ground
3.3. RangeRange [Ans. 2.0 s, hmax= 11.2 m, R = 17.4 m]
G r o u n d L e v e l
v 1
R a n g eh
U p
h M a x](https://image.slidesharecdn.com/u1c-m-l4-pm-in-vert-comp-academic-r-161022010943/85/Grade-11-U1C-L4-Vertical-PM-Academic-6-320.jpg)
Grade 11, U1C-L4, Vertical PM-Academic
- 1. Vertical Projectile Motion With An Initial Vertical Velocity Component Modified Lesson Nelson TB Reference Pages:Nelson TB Reference Pages: 78 (1/3 down) - 81 (1/3 down)78 (1/3 down) - 81 (1/3 down)
- 2. If an arrow, or any other object, is shot with anIf an arrow, or any other object, is shot with an angle between 0angle between 000 and 90and 9000 to the horizontal, theto the horizontal, the arrow (or object) will have an initial verticalarrow (or object) will have an initial vertical velocity component. For these types of problems,velocity component. For these types of problems, we must use the initial vertical velocitywe must use the initial vertical velocity component (vcomponent (v11sinsinθ)θ).. v H o r i z o n t a l = v 1 c o s ( ) vVertical=v1sin() v 1 U p
- 3. Case 1Case 1: An arrow is shot from ground level and: An arrow is shot from ground level and returns to ground level. This is a symmetricalreturns to ground level. This is a symmetrical trajectory.trajectory. Step 1: Find the time of flightStep 1: Find the time of flight ΔdΔd verticalvertical = v= v11 sin(θ)Δt + ½ a(Δt)sin(θ)Δt + ½ a(Δt)22 , but, but ΔdΔd verticalvertical = 0= 0,, andand a = - ga = - g So,So, Δt ( vΔt ( v11 sin(θ)sin(θ)- ½ g Δt) = 0- ½ g Δt) = 0,, after factoring out Δtafter factoring out Δt Thus, Δt = 0 orThus, Δt = 0 or Δt = (2vΔt = (2v11 sinθ)/gsinθ)/g,, we ignore Δt = 0we ignore Δt = 0 G r o u n d L e v e l U p R a n g e v 1
- 4. Chalkboard Example: Use the diagram from the previous page Let v1 = 100.0 m/s [300 up from horizontal] Determine: a.) time of fight b.) range c.)a.) time of fight b.) range c.) max height of projectile from ground d.)max height of projectile from ground d.) velocity of projectile just before it hits thevelocity of projectile just before it hits the groundground ANS a.) 10.2 s b.) 884 m c.) 128 ma.) 10.2 s b.) 884 m c.) 128 m d.) Discussd.) Discuss It is important to know that the greatestIt is important to know that the greatest range is obtained whenrange is obtained when θ = 450 andand greatest time of flight whengreatest time of flight when θ = 900 ..
- 5. Case 2Case 2 – Not a Symmetrical Trajectory– Not a Symmetrical Trajectory In the diagram, we see that the arrow achieves a finalIn the diagram, we see that the arrow achieves a final height of “h”. When we calculate the time of flight, weheight of “h”. When we calculate the time of flight, we would get two different and correct values.would get two different and correct values. One valuevalue would represent the arrow reaching height “h” on the waywould represent the arrow reaching height “h” on the way up ( at “A”). Theup ( at “A”). The second time value occurs when thetime value occurs when the arrow reaches the same height on the way down (at “B”).arrow reaches the same height on the way down (at “B”). To know which value is correct, more information wouldTo know which value is correct, more information would have to be given (range, achieving max height, etc.).have to be given (range, achieving max height, etc.). To find time of flight, in this example, a quadratic equationTo find time of flight, in this example, a quadratic equation would have to be solved since hwould have to be solved since h ≠≠ 00 ((h = 0 only for a symmetrical trajectoryh = 0 only for a symmetrical trajectory).). R a n g e v1 h A B
- 6. Chalkboard Example An arrow is shot with an initial velocity ofAn arrow is shot with an initial velocity of 10.0 m/s [300 Up from Horiz.] from a bridge which isfrom a bridge which is 10.0 m above the ground. Lettingabove the ground. Letting g = 10.0 m/s2 ,, determine the following:determine the following: 1.1. Time of flightTime of flight 2.2. Maximum height above the groundMaximum height above the ground 3.3. RangeRange [Ans. 2.0 s, hmax= 11.2 m, R = 17.4 m] G r o u n d L e v e l v 1 R a n g eh U p h M a x
- 7. Videos Very short http://www.youtube.com/watch?v=ZBfy-MNgtoYhttp://www.youtube.com/watch?v=ZBfy-MNgtoY Shooting at a toy monkey which drops from a tree – where do you aim? http://www.youtube.com/watch?v=cxvsHNRXLjwhttp://www.youtube.com/watch?v=cxvsHNRXLjw Long lesson-type video not shown in class. http://www.youtube.com/watch?v=-uUsUaPJUc0http://www.youtube.com/watch?v=-uUsUaPJUc0 Physics PM Rap! http://www.youtube.com/watch?v=3X5-nsbPz88http://www.youtube.com/watch?v=3X5-nsbPz88
- 8. Practice Problems Nelson TB: Page 81 #3, 4 (assume no air resistance), 6, 7, 8 Page 93 #41 {Consider ground level as being the height at which the ball leaves the barrel.} For additional practice, students should try the following McGraw-Hill problems found in the workbook Page 39B (New #42): #9-12#9-12