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4.5 First-order linear equations  (Page 4/10)

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Free fall with air resistance

We discussed air resistance at the beginning of this section. The next example shows how to apply this concept for a ball in vertical motion. Other factors can affect the force of air resistance, such as the size and shape of the object, but we ignore them here.

A ball with air resistance

A racquetball is hit straight upward with an initial velocity of 2 m/s. The mass of a racquetball is approximately 0.0427 kg. Air resistance acts on the ball with a force numerically equal to 0.5 v , where v represents the velocity of the ball at time t .

  1. Find the velocity of the ball as a function of time.
  2. How long does it take for the ball to reach its maximum height?
  3. If the ball is hit from an initial height of 1 meter, how high will it reach?
  1. The mass m = 0.0427 kg , k = 0.5 , and g = 9.8 m/s 2 . The initial velocity is v 0 = 2 m/s. Therefore the initial-value problem is
    0.0427 d v d t = −0.5 v − 0.0427 ( 9.8 ) , v 0 = 2 .

    Dividing the differential equation by 0.0427 gives
    d v d t = −11.7096 v − 9.8 , v 0 = 2 .

    The differential equation is linear. Using the problem-solving strategy for linear differential equations:
    Step 1. Rewrite the differential equation as d v d t + 11.7096 v = −9.8 . This gives p ( t ) = 11.7096 and q ( t ) = −9.8
    Step 2. The integrating factor is μ ( t ) = e ∫ 11.7096 d t = e 11.7096 t .
    Step 3. Multiply the differential equation by μ ( t ) :
    e 11.7096 t d v d t + 11.7096 v e 11.7096 t = −9.8 e 11.7096 t d d t [ v e 11.7096 t ] = −9.8 e 11.7096 t .

    Step 4. Integrate both sides:
    ∫ d d t [ v e 11.7096 t ] d t = ∫ −9.8 e 11.7096 t d t v e 11.7096 t = −9.8 11.7096 e 11.7096 t + C v ( t ) = −0.8369 + C e −11.7096 t .

    Step 5. Solve for C using the initial condition v 0 = v ( 0 ) = 2 :
    v ( t ) = −0.8369 + C e −11.7096 t v ( 0 ) = −0.8369 + C e −11.7096 ( 0 ) 2 = −0.8369 + C C = 2.8369.

    Therefore the solution to the initial-value problem is v ( t ) = 2.8369 e −11.7096 t − 0.8369 .
  2. The ball reaches its maximum height when the velocity is equal to zero. The reason is that when the velocity is positive, it is rising, and when it is negative, it is falling. Therefore when it is zero, it is neither rising nor falling, and is at its maximum height:
    2.8369 e −11.7096 t − 0.8369 = 0 2.8369 e −11.7096 t = 0.8369 e −11.7096 t = 0.8369 2.8369 ≈ 0.295 ln e −11.7096 t = ln 0.295 ≈ − 1.221 −11.7096 t = −1.221 t ≈ 0.104.

    Therefore it takes approximately 0.104 second to reach maximum height.
  3. To find the height of the ball as a function of time, use the fact that the derivative of position is velocity, i.e., if h ( t ) represents the height at time t , then h ′ ( t ) = v ( t ) . Because we know v ( t ) and the initial height, we can form an initial-value problem:
    h ′ ( t ) = 2.8369 e −11.7096 t − 0.8369 , h ( 0 ) = 1 .

    Integrating both sides of the differential equation with respect to t gives
    ∫ h ′ ( t ) d t = ∫ 2.8369 e −11.7096 t − 0.8369 d t h ( t ) = − 2.8369 11.7096 e −11.7096 t − 0.8369 t + C h ( t ) = −0.2423 e −11.7096 t − 0.8369 t + C .

    Solve for C by using the initial condition:
    h ( t ) = −0.2423 e −11.7096 t − 0.8369 t + C h ( 0 ) = −0.2423 e −11.7096 ( 0 ) − 0.8369 ( 0 ) + C 1 = −0.2423 + C C = 1.2423.

    Therefore
    h ( t ) = −0.2423 e −11.7096 t − 0.8369 t + 1.2423 .

    After 0.104 second, the height is given by
    h ( 0.2 ) = −0.2423 e −11.7096 t − 0.8369 t + 1.2423 ≈ 1.0836 meter.
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The weight of a penny is 2.5 grams (United States Mint, “Coin Specifications,” accessed April 9, 2015, http://www.usmint.gov/about_the_mint/?action=coin_specifications), and the upper observation deck of the Empire State Building is 369 meters above the street. Since the penny is a small and relatively smooth object, air resistance acting on the penny is actually quite small. We assume the air resistance is numerically equal to 0.0025 v . Furthermore, the penny is dropped with no initial velocity imparted to it.

  1. Set up an initial-value problem that represents the falling penny.
  2. Solve the problem for v ( t ) .
  3. What is the terminal velocity of the penny (i.e., calculate the limit of the velocity as t approaches infinity)?
  1. d v d t = − v − 9.8 v ( 0 ) = 0
  2. v ( t ) = 9.8 ( e − t − 1 )
  3. lim t → ∞ v ( t ) = lim t → ∞ ( 9.8 ( e − t − 1 ) ) = −9.8 m/s ≈ − 21.922 mph
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Source:  OpenStax, Calculus volume 2. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11965/1.2
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