Homework Solutions:pg. 41 # 16, 17, 18, pg. 43 # 36-             39, 41
Homework:                         pg. 41 # 16, 17, 18, pg. 43 # 36-39, 41  16. How would you estimate the maximum height y...
Homework:      pg. 41 # 16, 17, 18, pg. 43 # 36-39, 4136. (a) x = 31.9 m    (b) t = 5.1 s37. t = 1.5 seconds38. Vi = 16.2 ...
Homework:                    pg. 41 # 16, 17, 18, pg. 43 # 36-39, 41   36. A foul ball is hit straight up into the air wit...
Homework:                    pg. 41 # 16, 17, 18, pg. 43 # 36-39, 4137. A kangaroo jumps to a vertical height of 2.7 m. Ho...
Homework:                      pg. 41 # 16, 17, 18, pg. 43 # 36-39, 4138. A ball player catches a ball 3.3 seconds after t...
Homework:                  pg. 41 # 16, 17, 18, pg. 43 # 36-39, 41  39. Graph of a free-falling objectIf you designate dow...
Homework:                         pg. 41 # 16, 17, 18, pg. 43 # 36-39, 4141. A helicopter is ascending vertically with a s...
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H onors hw solutions on free fall

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H onors hw solutions on free fall

  1. 1. Homework Solutions:pg. 41 # 16, 17, 18, pg. 43 # 36- 39, 41
  2. 2. Homework: pg. 41 # 16, 17, 18, pg. 43 # 36-39, 41 16. How would you estimate the maximum height you could throw a ball vertically upward? How would you estimate the maximum speed you could- give it? time how long the ball is in the air until it returns back to you. You could- Divide this time by 2 (time to max height)- We know that velocity at max height will be zero- Can use the equations xf = 0.5at2 and vf = vi + at 17. An object that is thrown vertically upward will return to its original position with the same speed as it had initially if air resistance is negligible. If air resistance is appreciable,it willthis result be altered,speed if so, how? - Yes, it will be altered – will be less than the original and that it was thrown at 18. Describe in words the motion plotted in the figure in terms of v and a.- t = 0 s to t = 17 s: moving away from origin, constant velocity, no acceleration- t = 17 to t = 28 s: moving away from origin, increasing velocity, + acceleration- t = 28 s to t = 37 s: moving away from origin, decreasing velocity, - acceleration- t = 37 s to t = 45 s: moving towards origin, increasing negative velocity, - acceleration- t = 45 s to t = 50 s: moving towards origin, decreasing negative velocity, + acceleration
  3. 3. Homework: pg. 41 # 16, 17, 18, pg. 43 # 36-39, 4136. (a) x = 31.9 m (b) t = 5.1 s37. t = 1.5 seconds38. Vi = 16.2 m/s x = 13.4 m39. Graphs drawn41. t = 5.22 sec
  4. 4. Homework: pg. 41 # 16, 17, 18, pg. 43 # 36-39, 41 36. A foul ball is hit straight up into the air with a speed of about 25 m/s. (a) how high does it go? (b) How long is it in theXi = 0 m vf 2= vi2+ 2aΔx air? Xi = 0 m v = v + at f iXf = ? Xf =Vi = 25 m/s 0 = 252 + 2(-9.81)(xf – Vi = 25 m/s -25 = 25 + (-9.81)tVf = 0 m/s 0) Vf =-25 mm/sa = 9.81 m/s2 a = 9.81 m/s2 -50 = (-9.81)tt 0 = 625 + (-19.6)xf t=? t = 5.1 s -625 = (-19.6)xf Xf = 31.9 m
  5. 5. Homework: pg. 41 # 16, 17, 18, pg. 43 # 36-39, 4137. A kangaroo jumps to a vertical height of 2.7 m. How long was it in the air beforereturning to Earth? 1. Find vf with information about maximum heightXi = 0 mXf = 2.7 m vf 2= vi2+ 2aΔxVi = ? 0 = vi2 + 2(-9.81)(2.7)Vf = ?a = -9.81 m/s2 Vi2 = 2(9.81)(2.7)t Vi2 = √(52.9) Vi = 7.3 m/s 2. Solve for t using vf = vi + at using this vi, and a vf = -7.3 m/s -7.3 = 7.3 + (-9.81)t t = 1.5 s
  6. 6. Homework: pg. 41 # 16, 17, 18, pg. 43 # 36-39, 4138. A ball player catches a ball 3.3 seconds after throwing it vertically upward. Whatspeed did he throw it, and what is the maximum height? 1. Find viXi = 0 m xf = xi + vit +0.5at2Xf = 0 mVi = ? 0 = 0 + vi(3.3) +0.5(-9.81)(3.3)2Vf =a = -9.81 m/s2 Vi = 16.2 m/st = 3.3 s 2. Maximum height: use Vi = 0 m/s, and solve for xf vf 2= vi2+ 2aΔx 0 = (16.2)2 + 2(-9.81)(xf) xf = 13.4 m
  7. 7. Homework: pg. 41 # 16, 17, 18, pg. 43 # 36-39, 41 39. Graph of a free-falling objectIf you designate downward aspositive…
  8. 8. Homework: pg. 41 # 16, 17, 18, pg. 43 # 36-39, 4141. A helicopter is ascending vertically with a speed of 5.50 m/s. At a height of 105 m above the Earth, apackage is dropped from a window. How much time does it take for the package to reach the ground? 1. Find Vf =Xi = 0 mXf = -105 m vf 2= vi2+ 2aΔxVi = +5.50 m/sVf = vf 2= (5.50)2 + 2(-9.81)(-105)a = -9.81 m/s2t=? vf = -45.72 m/s OR 45.72 m/s in downward direction 1. Solve for t vf = vi + at -45.72 = +5.50 + (-9.81)t t = 5.22 s

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