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9/13 Projectile Motion  Today: examples  HW “9/13 Building” Due Wednesday, 9/18 On web or in 213 Witmer for copying  Last years practice exam on web.

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Presentation on theme: "9/13 Projectile Motion  Today: examples  HW “9/13 Building” Due Wednesday, 9/18 On web or in 213 Witmer for copying  Last years practice exam on web."— Presentation transcript:

1 9/13 Projectile Motion  Today: examples  HW “9/13 Building” Due Wednesday, 9/18 On web or in 213 Witmer for copying  Last years practice exam on web and in 213  Exam 1 Thursday 9/ pm Wit 116 and 114 me with conflicts and reasons to set up an alternative time.

2 Projectile Motion Example: An object is thrown with a velocity of 4m/s up and 3m/s west. The acceleration is 10m/s 2 directed down. Find the velocity at the point shown, 0/.4m above the ground. Last example we knew the final velocity by symmetry. Now we can’t. Split the motion into two parts, before the highest point and after the highest point. 0.4m tangent The velocity at the top is the x component.

3 Projectile Motion Example: An object is thrown with a velocity of 4m/s up and 3m/s west. The acceleration is 10m/s 2 directed down. Find the velocity at the point shown, 0.4m above the ground. 0.4m tangent y components: v y,f = 0m/s v y,i = 4m/s a y = 10m/s/s So  t = 0.4s  y = 0.8m y components: v y,i = 0m/s  y = 0.4m a y = 10m/s/s v y,f = ?m/s  t = 0.28s  v y = 2.8m/s v y,f = 2.8m/s v ave,y = 2m/s  y = 0.8m

4 Projectile Motion Example: An object is thrown with a velocity of 4m/s up and 3m/s west. The acceleration is 10m/s 2 directed down. Find the velocity at the point shown, 0.4m above the ground. 0.4m tangent y components: v y,0 = 0m/s v y,f = ?m/s  y = 0.4m a y = 10m/s/s  t = 0.28s  v y = 2.8m/s v y,f = 2.8m/s 2.8m/s 3.0m/s 4.1m/s   = 43   y = 0.8m  y = 0.4m

5 Projectile Motion Example: An object is thrown with an unknown velocity at 53  above the horizontal, reaching a height of 6.0m. The acceleration is 10m/s 2 directed down. Find the initial velocity. 6.0m


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