Amarble is dropped from a height of 1m a. how long will the ball be in the air before it strikes the ground? b. what was the average velocity of the ball during its flight c. how fast was the ball going the instant before it hit the ground

Step-by-step explanation:

a. We assume the appropriate equation for ballistic motion is ...

  h = -4.9t^2 +1

Then h = 0 when ...

  0 = -4.9t^2 +1

  49t^2 = 10 . . . . . add 4.9t^2, multply by 10

  7t = √10 . . . . . . . take the square root

  t = (√10)/7 . . . . . . divide by the coefficient of t

The marble will be in the air about (√10)/7 ≈ 0.451754 seconds.

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b. The average velocity is the ratio of distance to time:

  v = (1 m)/((√10)/7 s) = 0.7√10 m/s ≈ 2.214 m/s

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c. Under the influence of gravity, the velocity is linearly increasing over the time period, so its instantaneous value when the marble hits the ground will be twice the average value:

  When it hits, the marble's velocity is 1.4√10 m/s ≈ 4.427 m/s.

answer: 2530

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