Question

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.

Answer 1

each baloon hove 380 gram lauched with velocity that manybe 46.7 metre per second

Answer 2

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)