anonymous
  • anonymous
Suppose that a ball is dropped from the upper observation deck of the CN tower, 450 m above the ground. a) What is the velocity of the ball after 5 seconds? b)How fast is the ball traveling when it hits the ground? Thanks ! I really appreciate your help :)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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primeralph
  • primeralph
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Jhannybean
  • Jhannybean
a) you're measuring the vertical height of the ball by calculating the velocity of it's drop after 5 seconds. and because it's falling gravity is acting on it, thus (9.81 m\s^2 )(5 s) = ____ m/s b) s(t) = -4.9 * t^2 + 450 ft where s(t) is position. As it hits the ground, -4.9t^2 = 450 t = sqrt{ (450)/4.9 } t approx = 9.5831 s to calculate the velocity of its fall, take the derivative of the position function. s'(t) = -9.8t s'(t)= v(t) = -9.8 m\s^2 * (9.5831 s) v{t} = ? (units are in m/s)

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