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if an object is thrown up and lands in the same spot, which time is greater the ascent or descent, including air resistance. Why?
3 Answers
 Andrew SmithLv 74 weeks agoFavorite Answer
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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 oldprofLv 74 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 SmithLv 74 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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 billrussell42Lv 74 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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umm, good poijnt, I must have been a little punchy at the end.
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