Question

(1 point) A water balloon of mass 380 grams is launched with an initial (horizontal) velocity of 43 meters per second. As it travels, water leaks from the balloon at a rate of 60 grams per second. Assume air resistance is proportional to velocity with coefficient 5 grams per second. Again, because the motion is horizontal, ignore any effect due to gravity (a) Find the velocity of the balloon as a function of time. v(t) (b) What is the furthest distance the balloon could travel? Total distance = meters.

Answer 1

Fair Fair= kv5v

a(t)(t)
is acceleration then
F=ma=--5u
(as it is a resistive force) and
v(0)=43

So (ignoring any gravity effects)

1 380 5 76 380 LO

Meaning

In(u) In(C) (t) Cexp 76 76

And intial condition implies is the required velocity as a function of time

t v(t) 43 exp 76

t = / v(t) dt = D - 76 x 43 exp ( s(t) 76 r

t s(t)D 3268 exp 76
with
s(0)O

So we have

t s(t) 3268 3268 exp 76

Meaning the furthest distance it can get to is ${\color{Red} 3268}$ meters