# Homework Help: DE problems help

1. Apr 15, 2013

### uknowwho

1. The problem statement, all variables and given/known data

. A balloon is rising at the constant rate of 10 feet/second and is 100 feet from the ground at the instant when the astronaut drops his binoculars. (a) How long will it take the binoculars to strike the ground? (b) With what speed will the binoculars strike the ground?

2. A projectile is fired vertically upward by a cannon with an initial velocity of vo meters per second.At what speed will the projectile be moving when it returns and strikes the hapless cannoneer(Neglect air resistance)

3. Consider the differential equation dy/dt =k(A-y)(B+y) for the formation of trypsin in the small intestine.Assuming that A>B determing the time t at which trypsin is being formed most rapidly.

2. Relevant equations

3. The attempt at a solution

for (1)
a=-g
v=-gt + c1

v=0 and t=0

0=0 +c1
so c1=0

v=-gt

s=-gt^2/2 +c2

s=100 t=0

so 100=c2

s=-gt^2/2 + 100
s=-4.9t^2 + 100

putting s=0

-4.9t^2+100=0

getting t=4.5sec

and v=-44feet/sec

but the answer given at the back is t=2.83sec and v=80.62ft/sec

what am i doing wrong
When binoculars will be dropped at that time v will be zero and so will be the time and at the same time its distance from the ground will be 100feet
vi=v0 m/sec

for (2)

acc=-g
v=-gt+c1
v0=c1
v=-gt+v0

s=-gt^2/2 + v0t + c2

s=0 and t=0

c2=0

s=-gt^2/2 +vot

s=-4.9t^2 + vot

how to solve it further?

for (3)

dy/dt=k(A-y)(B+y)

dy/dt+k(AB+Ay-By-y^2)

dy/dt=ABk +Aky_Bky-ky^2
dy/dt=ABk + k(A-B)y-ky^2

dy/k(A-B)y-ky^2=(ABk) dt

Im stuck here

2. Apr 15, 2013

### Staff: Mentor

An astronaut in a balloon in a height of 100ft? Well, whatever.
The initial velocity of the binoculars is not 0 as the balloon is rising, and you are mixing feet and meters (in g) here. This would be obvious if you would work with units, which is a good idea in physics.

(2) is easy if you consider the symmetry of the setup, or use energy conservation. If you want to go the long way, set s=0 and solve for t. v follows as a result of t.

In (3), it is impossible to calculate t without initial conditions. It is possible to calculate y where the change of y reaches its maximum.

3. Apr 15, 2013

### uknowwho

for (1) if i put v=10,t=0 and s=100,t=0
i get v=-gt+10

and s=-4.9t^2+10t+100

-4.9t^2+10t+100=0

i get t as 5.65s

these should be the intial cdonditions?

for (3) how to calculate y where the change of y is max?
can you be more precise

4. Apr 15, 2013

### Staff: Mentor

I agree with your formulas for (1), but you are still mixing feet and meters.

How can you calculate the point where something is maximal in general? Just use that.

5. Apr 15, 2013

### SteamKing

Staff Emeritus
Do you know what g is in units of ft/s^2?

6. Apr 16, 2013

### uknowwho

I solved the other two but (3) i'm still not able to do..Can you just show the starting steps to give me an idea? it would be better if you could solve it so that I can understand

7. Apr 16, 2013

### Staff: Mentor

I am sure you calculated the maximum of functions before. How did you do it? Just do the same here.

I don't think it would be useful if I solve it. It is against the forum rules, and you can find many solved examples of similar problems anyway.
Did you have a look at the derivative?

8. Apr 16, 2013

### vela

Staff Emeritus
So you're saying you want to find where the function dy/dt attains a maximum. As mfb has suggested, that should ring some bells.