1 point) a water balloon of mass 380 grams is launched with an initial (vertical) velocity of 43 meters per second. Assume air resistance is proportional to velocity with coefficient 5 grams per second, and use 9.81 meters per second squared for the acceleration due to gravity.
Answers
each baloon hove 380 gram lauched with velocity that manybe 46.7 metre per second
Given info : a water balloon of mass 380 grams is launched with an initial (vertical) velocity of 43 meters per second. Assume air resistance is proportional to velocity with coefficient 5 grams per second.
To find : the velocity of the balloon as a function of time is..
solution : air resistance just opposes the relative motion of object.
i.e., air resistance acts downward in the above case.
now, net force = upward - downward
⇒ma = mg - air resistance
⇒a = g - air resistance/m
here, air resistance = 5v N
⇒a = 9.81 - 5v/380
⇒dv/dt = 9.81 - v/76
⇒76 dv/dt = 76 × 9.81 - v
⇒76 ∫dv/(745.56 - v) = ∫dt
⇒-76 [ln(745.56 - v)] = [t]
lower limit of v = 43 m/s
upper limit of v = v
for time range is 0 to t
so, 76 [ln(745.56 - 43) - ln(745.56 - v)] = t
⇒76 [ 6.55 - ln(745.56 - v) = t
⇒76 × 6.55 - 76 ln(745.56 - v) = t
⇒497.8 - 76 ln(745.56 - v) = t
⇒-76 ln(745.56 - v) = t - 497.8
⇒-ln (745.56 - v) = t/76 - 6.55
⇒745.56 - v = e^(-t/76 + 6.55)
⇒v = 745.56 - e^(-t/76 + 6.55) = 745.56 - 699.24e^(-t/76)
Therefore the velocity of the balloon as a function of time is 745.56 - 699.24e^(-t/76)