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dmitriy555 [2]
2 years ago
8

A hot-air balloon is ascending at the rate of 10 m/s and is 74 m above the ground when a package is dropped over the side. (a) H

ow long does the package take to reach the ground? (b) With what speed does it hit the ground?
Physics
1 answer:
Reika [66]2 years ago
4 0

Answer:

The answer to your question is:

a)  t1 = 2.99 s ≈ 3 s

b)  vf = 39.43 m/s

Explanation:

Data

vo = 10 m/s

h = 74 m

g = 9.81 m/s

t = ?   time to reach the ground

vf = ?   final speed

a)    h = vot + (1/2)gt²

     74 = 10t + (1/2)9.81t²

     4.9t² + 10t -74 = 0                  solve by using quadratic formula

   

   t = (-b ± √ (b² -4ac) / 2a

   t = (-10 ± √ (10² -4(4.9(-74) / 2(4.9)

   t = (-10 ± √ 1550.4 ) / 9.81

  t1 = (-10 + √ 1550.4 ) / 9.81               t2 = (-10 - √ 1550.4 ) / 9.81

  t1 = (-10 ± 39.38 ) / 9.81                    t2 = (-10 - 39.38) / 9.81

   t1 = 2.99 s ≈ 3 s                               t2 = is negative then is wrong there are

                                                                   no negative times.

b) Formula vf = vo + gt

                  vf = 10 + (9.81)(3)

                  vf = 10 + 29.43

                  vf = 39.43 m/s

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To solve this problem it is necessary to apply the concepts related to the flow as a function of the volume in a certain time, as well as the potential and kinetic energy that act on the pump and the fluid.

The work done would be defined as

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Where,

PE = Potential Energy

KE = Kinetic Energy

\Delta W = (\Delta m)gh+\frac{1}{2}(\Delta m)v^2

Where,

m = Mass

g = Gravitational energy

h = Height

v = Velocity

Considering power as the change of energy as a function of time we will then have to

P = \frac{\Delta W}{\Delta t}

P = \frac{\Delta m}{\Delta t}(gh+\frac{1}{2}v^2)

The rate of mass flow is,

\frac{\Delta m}{\Delta t} = \rho_w Av

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Final the power of the pump would be,

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