1) Given: vf = vi + at vi = 0 m/s vf = m/s t = 6 s a = -9.8 m/s2

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In this section you will:
vf2 = vi2 + 2a ∆x vf2 - vi2 2a (0 m/s)2 – (26.8 m/s)2 = ∆x 2(-3m/s2)
Presentation transcript:

1) Given: vf = vi + at vi = 0 m/s vf = -58.8 m/s t = 6 s a = -9.8 m/s2 58.8 m/s down 2) Given: vi = 0 m/s d = 0(3) + ½(-9.8)(3)2 d = vit +½at2 t = 3 s d = -44.1 m a = -9.8 m/s2; -32ft/s2 d = ? m redo with a = -32 ft/s2 If falls 44.1 m or 144 ft

tower is 219 ft tall; ball is moving at 118.4 ft/s down vf = vi + at 3) Given: d = 0(3.7) + ½(-32)(3.7)2 d = vit +½at2 vi = 0 ft/s d = -219 ft t = 3.7 s a = -32 ft/s2 vf = vi + at d = ? ft vf = -118.4 ft/s vf = ? ft/s vf = 0 + (-32)(3.7) tower is 219 ft tall; ball is moving at 118.4 ft/s down vf = vi + at 4) Given: vf = -116 ft/s vi = -20 ft/s vf = -20 + (-32)(3) t = 3 s a = -32 ft/s2 d = vit +½at2 d = (-20)(3) + ½(-32)(3)2 vf = ? ft/s d = ? ft d = -204 ft ball is moving at 116 ft/s down and fell 204 ft

5) 45.2 ft/s down 6)12.2 s; 735 m 7) 172 ft; 3.3 s 8) 49 ft 2d = (vf + vi)t vf = vi + at d = vit +½at2 vf2 = vi2 + 2ad