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Non-Horizontal Projectiles

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

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Non-Horizontal Projectiles

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

  2. 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?

  3. 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.

  4. 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.

  5. 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? This is what the trajectory looks like V1 40o

  6. 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? Δdy V1 40o ΔdH A. Maximum height or Δdy B. How far (range) or ΔdH

  7. 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? Δdy V1 40o ΔdH A. Maximum height or Δdy B. How far (range) or ΔdH

  8. 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 Vy= 0 Δdy V1 40o ΔdH V1y = (sin40)(23.5 m/s) = 15.1 m/s [U] V1y V1 40o VH = (cos40)(23.5 m/s) = 18.0 m/s [F] VH

  9. 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 Let’s make a plan for the first quantity we’ll need to solve for V2y= 0 Maximum height or Δdy (requires vertical info) Δdy V1 40o ΔdH • Which kinematics equation includes all 4 of these variables?

  10. 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] V2y= 0 Maximum height or Δdy (requires vertical info) Δdy V1 40o ΔdH • 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 + 2gΔdy2g (2) (-9.8m/s2[D][U])

  11. 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 V2y= 0 B. How far (range) or ΔdHrequires horizontal info Δdy V1 40o ΔdH • Which kinematics equation will allow us to solve for time using the vertical info we have?

  12. 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 V2y= 0 B. How far (range) or ΔdHrequires horizontal info Δdy V1 40o ΔdH • Which kinematics equation will allow us to solve for time using the vertical info we have? • this will give us the time to go up • V2 = V1 + aΔt- 1 Solving for Δt= V2y - V1y = 0 – 15.1 m/s [U] = 1.54 s • V2y = V1y + 2gΔt g - 9.8 m/s2 [D] [U]

  13. 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] V2y= 0 B. How far (range) or ΔdHrequires horizontal info Δdy V1 40o ΔdH

  14. 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 pageMaximum range ΔdH= V12sin2ΘwhereΘ = 45og 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 derive them 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! If you use any other angle, the object will not go as far. Equation will still calculate the range . V1 Θ ΔdH

  15. 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= V12sin2Θg = (23.5 m/s)2 (sin80) 9.8m/s = 55.5 m [F] Very simple!!!! V1 40o

  16. 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??? V1 Δdy ΔdH

  17. What do we know?Givens: V1 =23.5 m/s [@40o above horiz]Δdy= 3.00m • Solving for max height What will be different? V1 Δdy ΔdH

  18. What do we know?Givens: V1 =23.5 m/s [@40o above horiz]Δdy= 3.00m • Solving for max height What 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 V1 Δdup Max height Δdy ΔdH

  19. 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 V2y= 0 V1 Δdup Max height Δdy ΔdH Previous work gave us V1y = (sin40)(23.5 m/s) = 15.1 m/s [U] V2y = 0 g = 9.8 m/s2 [D]

  20. What do we know?Givens: V1 =23.5 m/s [@40o above horiz]Δdy= 3.00m B. Solving for total time of flight What will be different? V1 Δdy ΔdH

  21. What do we know?Givens: V1 =23.5 m/s [@40o above horiz]Δdy= 3.00m B. Solving for total time of flight What will be different? Our fancy equation won’t work because the object leaves and lands at different heights Back to kinematics! V1 Δdy ΔdH

  22. What do we know?Givens: V1 =23.5 m/s [@40o above horiz]Δdy= 3.00m Time to go up 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 V2y= 0; VH = 18.0 m/s V1 Time to go down Δdy ΔdH ΔtT = Δt up+ Δt down

  23. What do we know?Givens: V1 =23.5 m/s [@40o above horiz]Δdy= 3.00m Time to get back to the same height 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 V1 V1y= 15.1 m/s [D] Time to go below Δdy ΔdH This method requires you to use eq.3 when V1yis not 0; that means the quadratic equation!!!! ΔtT = Δt leave&land+ Δt down

  24. 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 directionGivens: 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 becauseΔt 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. V1y= 15.1 m/s [U] Δdy ΔdH Ax2 + Bx + C = 0 X= -b ±√b2-4ac 2a

  25. Givens: V1y =15.1 m/s [U] looking for Δt=???Δdy= 3.00m [D]g = 9.8 m/s2 [D] eq.3Δdy= V1y Δt +1/2g Δt2 we need to make this equation look more like the quadratic1/2g Δt2 + V1y Δt + -Δdy= 0Ax2 + Bx + C = 0½(9.8m/s2[D])+(-15.1 m/s[U][D])+-(3.00m[D])=0notice 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. V1y= 15.1 m/s [U] Δdy ΔdH Ax2 + Bx + C = 0 X= -b ±√b2-4ac 2a

  26. Givens: V1y =15.1 m/s [U] looking for Δt=???Δdy= 3.00m [D]g = 9.8 m/s2 [D] eq.3Δdy= V1y Δt +1/2g Δt2 Ax2+ Bx+ C = 01/2g Δt2 + V1y Δt + -Δdy= 0½(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. V1y= 15.1 m/s [U] Δdy ΔdH X= -b ±√b2-4ac 2a

  27. 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 Δt ΔdH = VHΔt = (18.0 m/s[F]) (3.27s) = 58.9 m [F] V1 Δdy ΔdH

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