Anonymous
Anonymous asked in Science & MathematicsPhysics · 4 weeks ago

if an object is thrown up and lands in the same spot, which time is greater the ascent or descent, including air resistance. Why?

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  • 4 weeks ago
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    The AVERAGE SPEED is higher on the way up.  The acceleration ( deceleration ) is higher but the final speed at the top is zero so the starting speed is higher than the landing speed.  As the speed is higher then the time for ascent is lower.  You can do this experiment yourself.

    To give more of the evidence.  Imagine that there is some friction force k(v)  which may or may not depend on v ( typically it is the form of F = b + cv + d v^2)  At some distance below the top of the flight h it has kinetic energy = 1/2 m v^2

    On the way up it loses energy = sum( k(v) dh)  ie mgh = 1/2 m v^2 - sum( k(v) dh)   On the way down it also loses energy  ( friction always absorbs energy)

    so 1/2 m v2 ^2 = mgh - sum(k(v) dh)

    -> 1/2 m v2^2 = 1/2 m v^2 - sum ( k(v) dh) -> v2 <v1 at each and every distance from the top "h". As v2<v1 at every single point in its flight it must take more time going down with this lower v2 than it did going up with a higher value of v

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  • 4 weeks ago

    My computer program shows that without drag the projectile will be in the air longer.  Mainly because it'll reach a higher maximum height.  See the output next:

    The experiment is defined by:

    Enter gravity accel. g = 9.81 m/sec^2

    Enter launch height = 0.000 meters above ground

     -

    Enter impact elevation = 0.000 meters ground level

    Initial launch height = 0.000 meters above impact

    Enter launch speed = 1000.000 mps

     -

    Initial launch speed U = 1000.000 mps 3,600.0 kph

    Enter launch angle = 90.00000 degrees 1.571 rads NOTE: straight up

    Uy = U sin(theta) = 1000.000 mps 3,600.0 kph

    Ux = U cos(theta) = 0.000 mps 0.0 kph

    Results Projectile in Air Trajectory

    NOTE: Enter drag characteristics here.

    Sea level atm. density rho = 1.2250 kg/m^3

    Coefficient of drag Cd = 0.0010

    Cross sectional area A = 0.0003 m^2

    Rest mass M = 0.008 kg

    y(max) = 50968.400 meters without drag <== NOTE: Without drag max height greater

    x(max) = 0.000 meters without drag

    T = 203.874 seconds without drag <== NOTE: About 30 seconds longer w/o drag

    y'(max) = 36944.227 meters with drag <== NOTE: Drag makes the max height less

    x'(max) = 0.000 meters with drag

    Flight time T = 172.005 seconds 2.867 minutes <== NOTE:  30 second shorter with drag

    Impact speed = 760.620 mps w/drag 2738.230 kph

    My colleague Andy pointed out I didn't really answer your question.  So I went back to the results and found this.  Without drag, the time up = time down = 101.5 seconds.  And with drag, time  up = 85 seconds and time down = 87 seconds.  ANS.

    This should be expected with the drag case because the object I used to do the experiment showed that the object during its fall never came close to the 1000 mps launch speed.  So the average speed falling was less than the average speed rising.

    • Andrew Smith
      Lv 7
      4 weeks agoReport

      That is not quite the question that was asked. Did it take more time on the way up or on the way down?

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  • 4 weeks ago

    No air resistance, they are the same.

    with air resistance:

    ascent, there are two forces pointing down, weight and air resistance. Taking air resistance as constant, those two forces add and produce an acceleration downward that is greater than g

    F = f+w where f is force of air resistance

    F = f+mg

    a₀ = F/m = f/m + mg/m = f/m + g

    and t₀ = v₀/a₀ = v₀/(g + f/m)

    where v₀ is the initial velocity

    and t₀ is less than the time without air resistance, 

    h₀ = v₀²/2a₀ = v₀²/2(g + f/m) 

    which is less than the height it would normally reach without air resistance

    descent, the force of weight, pointing down is opposed by the force of air resistance, so a₁ = g – f/m

    falling from that height

    t₁ = √(2h₀/a₁) = √(2(v₀²/2(g + f/m) )/a₁)

    t₁ = v₀√(1/a₁(g + f/m) )

    t₁ = v₀√(1/(g – f/m)(g + f/m))

    t₁ = v₀√(1/(g² – f²/m²))

    now that looks complicated, but compare to

    t₀ = v₀/(g + f/m)

    take some typical numbers

    v₀ = 100 m/s

    g = 10 m/s²

    m = 1 kg

    w = 10 N

    f = 1 N

    with no friction 

    t = v/g = 100/10 = 10 s

    t₀ = v₀/(g + f/m) = 100/(10+1) = 9 s

    t₁ = v₀√(1/(g² – f²/m²)) = 100√(1/(100 – 1)) = 10.05 s

    so as expected, slower going up, faster going down.

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