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M3 linear motion with variable forces

this actually seems like an easy question-
A bullet of mass 50 grams fired vertically upwards at speed 600m/s,at a speed v it experiences air resistance kv^2,where k=4.9x10^-5, find the time taken by the bullet to reach its max height
what ive done so far
mdvdt=−kv2−mg m\frac{dv}{dt}=-kv^2-mg
dtdv=m−kv2−mg\frac{ dt }{ dv } = \frac{m}{-kv^2 -mg}
what i was planning to do next is to integrate the right part of the bottom equation,and make it opposite signs and put in limits 600 and 0, my problem was i dont know how to integrate that last part, or have i gone wrong?
(edited 10 years ago)

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Reply 1
Avatar for davros
davros
16
Original post by physics4ever
this actually seems like an easy question-
A bullet of mass 50 grams fired vertically upwards at speed 600m/s,at a speed v it experiences air resistance kv^2,where k=4.9x10^-5, find the time taken by the bullet to reach its max height
what ive done so far
mdvdt=−kv2−mg m\frac{dv}{dt}=-kv^2-mg
dtdv=m−kv2−mg\frac{ dt }{ dv } = \frac{m}{-kv^2 -mg}
what i was planning to do next is to integrate the right part of the bottom equation,and make it opposite signs and put in limits 600 and 0, my problem was i dont know how to integrate that last part, or have i gone wrong?


I haven't checked your reasoning/working, but 1/(v^2 + a^2) can either be found with a trig substitution v = atan u or looked up from standard tables (see formula book!) :smile:
You want to integrate dv/dt, not dt/dv. Consider that when the bullet is at it's maximum height, v=0.

So we want to integrate to get an equation like v = (some function of t), then set it equal to zero.

Also, you shouldn't substitute in the initial velocity as a limit. Once you do the integration (indefinite integration), you'll use the v=600m/s at time t=0 (the initial conditions) to find the value of the "+c".
(edited 10 years ago)
Reply 3
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physics4ever
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Original post by dantheman1261

So we want to integrate to get an equation like v = (some function of t), then set it equal to zero.

how do i integrate dv/dt
Reply 4
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physics4ever
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Original post by davros
I haven't checked your reasoning/working, but 1/(v^2 + a^2) can either be found with a trig substitution v = atan u or looked up from standard tables (see formula book!) :smile:

the formula book doesnt have it in the form where theres a coefficient on v^2?
Davros has the correct method. Remember kv2=((k)v)2 k v^2 = ((\sqrt{k})v)^2
Original post by physics4ever
the formula book doesnt have it in the form where theres a coefficient on v^2?


Then take out a factor.
Reply 7
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physics4ever
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Original post by ghostwalker
Then take out a factor.

Original post by davros
I haven't checked your reasoning/working, but 1/(v^2 + a^2) can either be found with a trig substitution v = atan u or looked up from standard tables (see formula book!) :smile:

Original post by dantheman1261
Davros has the correct method. Remember kv2=((k)v)2 k v^2 = ((\sqrt{k})v)^2


am i suppose to find what a^2 is?because i cant
Original post by physics4ever
am i suppose to find what a^2 is?because i cant


Using a slightly different example.


1kv2+1=1k×1v2+1k\dfrac{1}{kv^2+1}= \dfrac{1}{k}\times\dfrac{1}{v^2+\frac{1}{k}}
Reply 9
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physics4ever
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Original post by ghostwalker
Using a slightly different example.


1kv2+1=1k×1v2+1k\dfrac{1}{kv^2+1}= \dfrac{1}{k}\times\dfrac{1}{v^2+\frac{1}{k}}

i did do that at the start and i got
m−kv2−mg=−mk×1v2+mgk\dfrac{m}{-kv^2-mg}= \dfrac{-m}{k}\times\dfrac{1}{v^2+\frac{mg}{k}}
and then i didnt know how to integrate this because theres a coefficient infront of 1/a^2+v^2
Reply 10
Avatar for davros
davros
16
Original post by physics4ever
i did do that at the start and i got
m−kv2−mg=−mk×1v2+mgk\dfrac{m}{-kv^2-mg}= \dfrac{-m}{k}\times\dfrac{1}{v^2+\frac{mg}{k}}
and then i didnt know how to integrate this because theres a coefficient infront of 1/a^2+v^2


Why should the coefficient give you a problem? Surely you're happy that

∫kf(x)dx=k∫f(x)dx\displaystyle \int kf(x) dx = k \int f(x) dx
Original post by physics4ever
i did do that at the start and i got
m−kv2−mg=−mk×1v2+mgk\dfrac{m}{-kv^2-mg}= \dfrac{-m}{k}\times\dfrac{1}{v^2+\frac{mg}{k}}
and then i didnt know how to integrate this because theres a coefficient infront of 1/a^2+v^2


That's just a constant, and you can pull it out in front of the integral.
Original post by ghostwalker
That's just a constant, and you can pull it out in front of the integral.

i tried doing that and then i got
t=[mk×1(mg/k)0.5×tan−1(v(mg/k)0.5)]0600t=\left[ \dfrac{m}{k}\times\dfrac{1}{(mg/k)^{0.5} }\times tan^{-1}(\frac{v}{ (mg/k)^{0.5}})\right]_{0}^{600} with a^2=mg/k,
but my answer was t=about 800 the answer in the back of my textbook was about 14 haha not even close
(edited 10 years ago)
Original post by physics4ever
i tried doing that and then i got
t=[mk×1(mg/k)0.5×tan−1(v(mg/k)0.5)]0600t=\left[ \dfrac{m}{k}\times\dfrac{1}{(mg/k)^{0.5} }\times tan^{-1}(\frac{v}{ (mg/k)^{0.5}})\right]_{0}^{600} with a^2=mg/k,
but my answer was t=about 800 the answer in the back of my textbook was about 14 haha not even close


Without checking your formula, my first thought is do you have everything in consistent units.
Original post by ghostwalker
Without checking your formula, my first thought is do you have everything in consistent units.

yes i have m=50x10^-3 k=4.9x10^-5

and i got my formula from the data sheet which is

∫(a2+x2)−1dx=(1/a)×tan⁡−1(x/a)\displaystyle\int (a^2+x^2)^{-1}dx=(1/a)\times\tan^{-1}(x/a)
(edited 10 years ago)
Original post by physics4ever
yes i have m=50x10^-3 k=4.9x10^-5

and i got my formula from the data sheet which is

∫(a2+x2)−1dx=(1/a)×tan⁡−1(x/a)\displaystyle\int (a^2+x^2)^{-1}dx=(1/a)\times\tan^{-1}(x/a)


Were you given any units for k ?
Original post by ghostwalker
Were you given any units for k ?

k was just a constant so there was no units for k,it just said k=4.9x10^-5
Original post by physics4ever
k was just a constant so there was no units for k,it just said k=4.9x10^-5


The other thing is did you have your calculator in radians?

As I got 14.34... when evaluating that formula.

As an interim result, I got root(mg/k) = 100
Original post by ghostwalker
The other thing is did you have your calculator in radians?

As I got 14.34... when evaluating that formula.

As an interim result, I got root(mg/k) = 100

i had mine in degrees, i didnt know it had to be in rads as it didnt say in the datasheet,
and ive never read in the core 3 or 4 textbooks how the integral of 1/(a^2+x^2) was derived so i knew nothing about it and just slotted numbers in, id like to know though
Original post by physics4ever
i had mine in degrees, i didnt know it had to be in rads as it didnt say in the datasheet,
and ive never read in the core 3 or 4 textbooks how the integral of 1/(a^2+x^2) was derived so i knew nothing about it and just slotted numbers in, id like to know though


Any calculus done with trig functions has the argument/result as appropriate in radians.

d(cos(x))/dx = - sin(x) is only true if x is in radians, etc.
(edited 10 years ago)

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