NonHorizontal Projectiles Last class you were given projectile motion notes and some sample...

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Transcript of NonHorizontal Projectiles Last class you were given projectile motion notes and some sample...
NonHorizontal Projectiles
NonHorizontal ProjectilesLast class you were given projectile motion notes and some sample problems. This is the nonhorizontal 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 nonhorizontal 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 nonhorizontal 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 subcategories that I refer to as leave and land at same height or leave and land belowthe 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?V140oThis is what the trajectory looks likeA 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?
V140oA. Maximum height or dyB. How far (range) or dHdHdyA 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?
V140oA. Maximum height or dyB. How far (range) or dHdHdyA 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 heightg = 9.8 m/s2[D]V1 can be broken into vertical & horizontal componentsV140odHdyVy= 0V1V1yVH40oV1y = (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
V140oMaximum height or dy (requires vertical info)Which kinematics equation includes all 4 of these variables?
dHdyV2y= 0Lets make a plan for the first quantity well need to solve forA 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]
V140oMaximum height or dy (requires vertical info)Which kinematics equation includes all 4 of these variables?
V2 2= V1 2 + 2ad  5Solving for dy = V2y 2 V1y 2 = (0) (15.1m/s [U])2V2y 2= V1y 2 + 2gdy 2g (2) (9.8m/s2[D][U])
dHdyV2y= 0A 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 speedMust use vH = dH we need t t
V140oWhich kinematics equation will allow us to solve for time using the vertical info we have?
dHdyV2y= 0B. How far (range) or dH requires horizontal infoA 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 speedMust use vH = dH we need t t
V140oWhich 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 + at  1Solving for t = V2y  V1y = 0 15.1 m/s [U] = 1.54 sV2y = V1y + 2gt g  9.8 m/s2 [D] [U]dHdyV2y= 0B. How far (range) or dH requires horizontal infoA 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] tup = 1.54s tT = (1.54s)(2)= 3.08sRearrange vH = dH to solve for dH tdH = vH t = (18.0 m/s [F]) (3.08 s) = 55.4 m [F]
V140odHdyV2y= 0B. How far (range) or dH requires horizontal infoThere 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 dH = V12 sin2where = 45o g 2= (2x45)=90 and sin90=1
Total time of flight t = 2V1 sin g
These equations come from kinematics,there are no tricks. You could even derivethem yourselves if youd 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 onestep equation!V1dHIf 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] dH = V12 sin2 g = (23.5 m/s)2 (sin80)9.8m/s= 55.5 m [F]
Very simple!!!!V140oWhat would be different in the Leave & Land Below situation?Lets 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???dydHV1
What do we know?Givens: V1 = 23.5 m/s [@40o above horiz] dy= 3.00m
Solving for max heightWhat will be different?dydHV1
What do we know?Givens: V1 = 23.5 m/s [@40o above horiz] dy= 3.00m
Solving for max heightWhat will be different?this is not just dy now; we have to find out how high the ball will go up and add the 2 values together
Max height = dy + dup dydHV1Max heightdup
What do we know?Givens: V1 = 23.5 m/s [@40o above horiz] dy= 3.00m
Max height = dy + dup = 3.00m + dup
V2y 2= V1y 2 + 2gdySolving for dup = V2y 2 V1y 2 = (0) (15.1m/s [U])22g(2) (9.8m/s2[D][U]) = 11.6 m
So max height = 3.00 + 11.6m = 14.6 m from the level green
dydHV1Max heightdupPrevious 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] dy= 3.00m
B. Solving for total time of flightWhat will be different?
dydHV1
What do we know?Givens: V1 = 23.5 m/s [@40o above horiz] dy= 3.00m
B. Solving for total time of flightWhat will be different?
Our fancy equation wont work because the object leaves and lands at different heights
Back to kinematics!
dydHV1
What do we know?Givens: V1 = 23.5 m/s [@40o above horiz] dy= 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
dydHV1Time to go upV2y= 0; VH = 18.0 m/sTime to go downtT = t up + t down
What do we know?Givens: V1 = 23.5 m/s [@40o above horiz] dy= 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
dydHV1Time to get back to the same heightTime to go belowV1y= 15.1 m/s [D]This method requires you to use eq.3 when V1yis not 0; that means the quadratic equation!!!!tT = t leave&land + t down
What do we know? Since we are looking for time we wont 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 t=???dy= 3.00m [D]g = 9.8 m/s2 [D]eq.3 dy= V1y t +1/2g t2
because t 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.
dydHV1y= 15.1 m/s [U]Ax2 + Bx + C = 0
X= b b24ac 2a
Givens: V1y = 15.1 m/s [U] looking for t=???dy= 3.00m [D]g = 9.8 m/s2 [D]eq.3 dy= V1y t +1/2g t2
we need to make this equation look more like the quadratic 1/2g t2 + V1y t + dy = 0
Ax2 + Bx + C = 0
(9.8m/s2[D])+(15.1 m/s[U][D])+(3.00m[D])=0
notice that Ive made all the vectors into [D] as the Aterm 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.
dydHV1y= 15.1 m/s [U]Ax2 + Bx + C = 0
X= b b24ac 2a
Givens: V1y = 15.1 m/s [U] looking for t=???dy= 3.00m [D]g = 9.8 m/s2 [D]eq.3 dy= V1y t +1/2g t2
Ax2 + Bx + C = 0 1/2g t2 + V1y t + dy = 0
(9.8m/s2[D])+(15.1 m/s[U][D])+(3.00m[D])=0 t = (15.1) (15.1)24(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.
dydHV1y= 15.1 m/s [U]X= b b24ac 2aC. 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 = dH tdH = VHt = (18.0 m/s[F]) (3.27s) = 58.9 m [F]dydHV1