non-horizontal projectiles last class you were given projectile motion notes and some sample...
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Non-Horizontal Projectiles
Last class you were given projectile motion notes and some sample problems. This is the non-horizontal problem......
A golfer hits a golf ball with an initial velocity of 23.5 m/s at an angle of 40o to the fairway towards a level green.A. What will be its maximum height?B. How far (horizontally) will the ball travel?
• What makes this a non-horizontal projectile problem?
A golfer hits a golf ball with an initial velocity of 23.5 m/s at an angle of 40o to the fairway towards a level green.A. What will be its maximum height?B. How far (horizontally) will the ball travel?
• Whenever the velocity is anything other than forwards (some angle given or implied), it falls into the non-horizontal category.
A golfer hits a golf ball with an initial velocity of 23.5 m/s at an angle of 40o to the fairway towards a level green.A. What will be its maximum height?B. How far (horizontally) will the ball travel?
• This category splits into 2 main sub-categories that I refer to as “leave and land at same height” or “leave and land below”
• the “level green” makes it clear that the object begins and ends its motion at the same height.
A golfer hits a golf ball with an initial velocity of 23.5 m/s at an angle of 40o to the fairway towards a level green.A. What will be its maximum height?B. How far (horizontally) will the ball travel?
Given Quantities:V1= 23.5 m/s [@40o above horiz]
What are we solving for?
V140o
This is what the trajectory looks like
A golfer hits a golf ball with an initial velocity of 23.5 m/s at an angle of 40o to the fairway towards a level green.A. What will be its maximum height?B. How far (horizontally) will the ball travel?
Given Quantities:V1= 23.5 m/s [@40o above horiz]
What are we solving for?
V140o
A. Maximum height or Δdy B. How far (range) or ΔdH
ΔdH
Δdy
A golfer hits a golf ball with an initial velocity of 23.5 m/s at an angle of 40o to the fairway towards a level green.A. What will be its maximum height?B. How far (horizontally) will the ball travel?
Given Quantities:V1= 23.5 m/s [@40o above horiz]
What else do we know?
What are we solving for?
V140o
A. Maximum height or Δdy B. How far (range) or ΔdH
ΔdH
Δdy
A golfer hits a golf ball with an initial velocity of 23.5 m/s at an angle of 40o to the fairway towards a level green.A. What will be its maximum height?B. How far (horizontally) will the ball travel?
Given Quantities: (both horizontal & vertical information)
V1= 23.5 m/s [@40o above horiz]
Vy= 0 at max height
g = 9.8 m/s2[D]V1 can be broken into vertical & horizontal components
V140o
ΔdH
Δdy
Vy= 0
V1V1y
VH
40o
V1y = (sin40)(23.5 m/s) = 15.1 m/s [U]
VH = (cos40)(23.5 m/s) = 18.0 m/s [F]
A golfer hits a golf ball with an initial velocity of 23.5 m/s at an angle of 40o to the fairway towards a level green.A. What will be its maximum height?B. How far (horizontally) will the ball travel?
Given Quantities:V1= 23.5 m/s [@40o above horiz]
We know: V1y = 15.1 m/s [U]
g = 9.8 m/s2[D]V2y= 0 at the max height
V140o
A. Maximum height or Δdy
(requires vertical info)
Which kinematics equation includes all 4 of these variables?
ΔdH
Δdy
V2y= 0
Let’s make a plan for the first quantity we’ll need to solve for
A golfer hits a golf ball with an initial velocity of 23.5 m/s at an angle of 40o to the fairway towards a level green.A. What will be its maximum height?B. How far (horizontally) will the ball travel?
Given Quantities:V1= 23.5 m/s [@40o above horiz]
We know: V1y = 15.1 m/s [U]
g = 9.8 m/s2[D]V2y= 0 at the max height
= 11.6 m [U]
V140o
A. Maximum height or Δdy
(requires vertical info)
Which kinematics equation includes all 4 of these variables?
V2 2= V1
2 + 2aΔd - 5 Solving for Δdy = V2y 2- V1y
2 = (0)- (15.1m/s [U])2
V2y 2= V1y
2 + 2gΔdy 2g (2) (-9.8m/s2[D][U])
ΔdH
Δdy
V2y= 0
A golfer hits a golf ball with an initial velocity of 23.5 m/s at an angle of 40o to the fairway towards a level green.A. What will be its maximum height?B. How far (horizontally) will the ball travel?
Given Quantities:V1= 23.5 m/s [@40o above horiz]
We know: VH = 18.0 m/s [F]
Motion in this direction happens at a constant speed
Must use vH = ΔdH we need Δt
Δt
V140o
Which kinematics equation will allow us to solve for time using the vertical info we have?
ΔdH
Δdy
V2y= 0 B. How far (range) or ΔdH requires horizontal info
A golfer hits a golf ball with an initial velocity of 23.5 m/s at an angle of 40o to the fairway towards a level green.A. What will be its maximum height?B. How far (horizontally) will the ball travel?
Given Quantities:V1= 23.5 m/s [@40o above horiz]
We know: VH = 18.0 m/s [F]
Motion in this direction happens at a constant speed
Must use vH = ΔdH we need Δt
Δt
V140o
Which kinematics equation will allow us to solve for time using the vertical info we have?
this will give us the time to go upV2 = V1
+ aΔt - 1 Solving for Δt = V2y - V1y = 0 – 15.1 m/s [U] = 1.54 sV2y = V1y
+ 2gΔt g - 9.8 m/s2 [D] [U]
ΔdH
Δdy
V2y= 0 B. How far (range) or ΔdH requires horizontal info
A golfer hits a golf ball with an initial velocity of 23.5 m/s at an angle of 40o to the fairway towards a level green.A. What will be its maximum height?B. How far (horizontally) will the ball travel?
Given Quantities:V1= 23.5 m/s [@40o above horiz]
So now we know: VH = 18.0 m/s [F]
Δtup = 1.54s ΔtT = (1.54s)(2)= 3.08s
Rearrange vH = ΔdH to solve for ΔdH
Δt
ΔdH = vH Δt
= (18.0 m/s [F]) (3.08 s) = 55.4 m [F]
V140o
ΔdH
Δdy
V2y= 0B. How far (range) or ΔdH requires horizontal info
There are 2 handy equations that are nice shortcuts for Leave & Land at the same height projectiles. You have been given them on your notes page
Maximum range ΔdH = V12
sin2Θ where Θ = 45o
g 2Θ= (2x45)=90 and sin90=1
Total time of flight Δt = 2V1 sinΘ g
These equations come from kinematics,there are no tricks. You could even derivethem yourselves if you’d like!
When you use them, just be careful to notice that you use V1 not a component. Try calculating the range of the ball using this one-step equation!
V1Θ
ΔdH
If you use any other angle, the object will not go as far. Equation will still calculate the range .
A golfer hits a golf ball with an initial velocity of 23.5 m/s at an angle of 40o to the fairway towards a level green.
B. How far (horizontally) will the ball travel?
Given Quantities:V1= 23.5 m/s [@40o above horiz]
ΔdH = V12
sin2Θ g
= (23.5 m/s)2 (sin80)9.8m/s
= 55.5 m [F]
Very simple!!!!
V140o
What would be different in the “Leave & Land Below” situation?
• Let’s take the same problem and add a twist...the golf tee is set above the level green by 3.00 m.
• How would this affect how we determine height , time of flight and the range of the ball???
Δdy
ΔdH
V1
What do we know?Givens: V1 = 23.5 m/s [@40o above horiz]
Δdy= 3.00m
A. Solving for max heightWhat will be different?
Δdy
ΔdH
V1
What do we know?Givens: V1 = 23.5 m/s [@40o above horiz]
Δdy= 3.00m
A. Solving for max heightWhat will be different?- this is not just Δdy now; we
have to find out how high the ball will go up and add the 2 values together
Max height = Δdy + Δdup
Δdy
ΔdH
V1
Max height
Δdup
What do we know?Givens: V1 = 23.5 m/s [@40o above horiz]
Δdy= 3.00m
Max height = Δdy + Δdup
= 3.00m + Δdup
V2y 2= V1y
2 + 2gΔdy
Solving for Δdup = V2y
2- V1y 2 = (0)- (15.1m/s [U])2
2g (2) (-9.8m/s2[D][U])
= 11.6 m
So max height = 3.00 + 11.6m = 14.6 m from the level green
Δdy
ΔdH
V1
Max height
Δdup
Previous work gave usV1y = (sin40)(23.5 m/s) = 15.1 m/s [U]V2y = 0g = 9.8 m/s2 [D]
V2y= 0
What do we know?Givens: V1 = 23.5 m/s [@40o above horiz]
Δdy= 3.00m
B. Solving for total time of flightWhat will be different?
Δdy
ΔdH
V1
What do we know?Givens: V1 = 23.5 m/s [@40o above horiz]
Δdy= 3.00m
B. Solving for total time of flightWhat will be different?
Our fancy equation won’t work because the object leaves and lands at different heights
Back to kinematics!
Δdy
ΔdH
V1
What do we know?Givens: V1 = 23.5 m/s [@40o above horiz]
Δdy= 3.00m
We could slice this diagram a number of ways to calculate the time
Option 1: solve for the time to go up, exactly the same way as before
and then treat the other section of the trajectory like a horizontal projectile and solve for the time to hit the ground
Δdy
ΔdH
V1
Time to go up
V2y= 0; VH = 18.0 m/s
Time to go down
ΔtT = Δt up + Δt down
What do we know?Givens: V1 = 23.5 m/s [@40o above horiz]
Δdy= 3.00m
We could slice this diagram a number of ways to calculate the time
Option 2: use our time of flight equation for the leave and land at same height and then use kinematics to find the time to fall the rest of the way to ground level
Δdy
ΔdH
V1
Time to get back to the same height
Time to go below
V1y= 15.1 m/s [D]
This method requires you to use eq.3 when V1yis not 0; that means the quadratic equation!!!!
ΔtT = Δt leave&land + Δt down
What do we know? Since we are looking for time we won’t have enough info to use the horizontal information If we think about what we know in the vertical direction
Givens: V1y = 15.1 m/s [U] looking for Δt=???Δdy= 3.00m [D]g = 9.8 m/s2 [D]
eq.3
Δdy= V1y Δt +1/2g Δt2
because Δt is in 2 different places in 2 different forms the only way to solve is the quadratic equation
Option 3: (best in my books)Requires the quadratic equation and does the whole thing in one process.
Δdy
ΔdH
V1y= 15.1 m/s [U]
Ax2 + Bx + C = 0
X= -b ±√b2-4ac 2a
Givens: V1y = 15.1 m/s [U] looking for Δt=???Δdy= 3.00m [D]g = 9.8 m/s2 [D]
eq.3 Δdy= V1y Δt +1/2g Δt2
we need to make this equation look more like the quadratic
1/2g Δt2 + V1y Δt + -Δdy = 0
Ax2 + Bx + C = 0
½(9.8m/s2[D])+(-15.1 m/s[U][D])+-(3.00m[D])=0
notice that I’ve made all the vectors into [D] as the A-term must be positive and to do vector math the directions must match
Option 3: (best in my books)Requires the quadratic equation and does the whole thing in one process.
Δdy
ΔdH
V1y= 15.1 m/s [U]
Ax2 + Bx + C = 0
X= -b ±√b2-4ac 2a
•
Givens: V1y = 15.1 m/s [U] looking for Δt=???Δdy= 3.00m [D]g = 9.8 m/s2 [D]
eq.3 Δdy= V1y Δt +1/2g Δt2
Ax2 + Bx + C = 0 1/2g Δt2 + V1y Δt + -Δdy = 0
½(9.8m/s2[D])+(-15.1 m/s[U][D])+-(3.00m[D])=0
Δt = -(-15.1) ± √(-15.1)2-4(4.9)(-3.00) 2(4.9) = +15.1 ± 16.9 since we are solving for time 9.8 only a positive answer makes sense = 3.27 s
Option 3: (best in my books)Requires the quadratic equation and does the whole thing in one process.
Δdy
ΔdH
V1y= 15.1 m/s [U]
X= -b ±√b2-4ac 2a
C. Solving for the range of the ball
Now that we have solved for time, we can very easily plug into the constant speed equation
VH = ΔdH
ΔtΔdH = VHΔt = (18.0 m/s[F]) (3.27s) = 58.9 m [F]
Δdy
ΔdH
V1