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......

Transcript of Non-Horizontal Projectiles Last class you were given projectile motion notes and some sample...

Page 1: Non-Horizontal Projectiles Last class you were given projectile motion notes and some sample problems. This is the non-horizontal problem......

Non-Horizontal Projectiles

Last class you were given projectile motion notes and some sample problems. This is the non-horizontal problem......

Page 2: 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?

Page 3: 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?

• Whenever the velocity is anything other than forwards (some angle given or implied), it falls into the non-horizontal category.

Page 4: 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?

• 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.

Page 5: 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?

Given Quantities:V1= 23.5 m/s [@40o above horiz]

What are we solving for?

V140o

This is what the trajectory looks like

Page 6: 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?

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

Page 7: 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?

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

Page 8: 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?

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]

Page 9: 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?

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

Page 10: 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?

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

Page 11: 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?

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

Page 12: 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?

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

Page 13: 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?

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

Page 14: Non-Horizontal Projectiles Last class you were given projectile motion notes and some sample problems. This is the non-horizontal problem......

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 .

Page 15: 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.

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

Page 16: Non-Horizontal Projectiles Last class you were given projectile motion notes and some sample problems. This is the non-horizontal problem......

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

Page 17: Non-Horizontal Projectiles Last class you were given projectile motion notes and some sample problems. This is the non-horizontal problem......

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

Page 18: Non-Horizontal Projectiles Last class you were given projectile motion notes and some sample problems. This is the non-horizontal problem......

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

Page 19: Non-Horizontal Projectiles Last class you were given projectile motion notes and some sample problems. This is the non-horizontal problem......

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

Page 20: Non-Horizontal Projectiles Last class you were given projectile motion notes and some sample problems. This is the non-horizontal problem......

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

Page 21: Non-Horizontal Projectiles Last class you were given projectile motion notes and some sample problems. This is the non-horizontal problem......

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

Page 22: Non-Horizontal Projectiles Last class you were given projectile motion notes and some sample problems. This is the non-horizontal problem......

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

Page 23: Non-Horizontal Projectiles Last class you were given projectile motion notes and some sample problems. This is the non-horizontal problem......

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

Page 24: Non-Horizontal Projectiles Last class you were given projectile motion notes and some sample problems. This is the non-horizontal problem......

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

Page 25: Non-Horizontal Projectiles Last class you were given projectile motion notes and some sample problems. This is the non-horizontal problem......

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

Page 26: Non-Horizontal Projectiles Last class you were given projectile motion notes and some sample problems. This is the non-horizontal problem......

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

Page 27: Non-Horizontal Projectiles Last class you were given projectile motion notes and some sample problems. This is the non-horizontal problem......

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