Two fledglings leave a nest which is 2.50 m off the ground. One drops from rest and then 0.10 s later the second charges out of the nest with a velocity with horizontal and downward components of 1.00 and 1.50 m/s, respectively.
1. Which fledgling hits the ground first (first or second)?
2. With what speed, in m/s, does the first fledgling hit the ground?
3. With what speed, in m/s, does the second fledgling hit the ground?
4. Which fledgling travels the greater displacement?
5. How far does a parent bird have to travel on the ground between the two fledglings, in m, to make sure they are alright?

t

1) the time of the pigeon 1 is less, so it comes first

2) v = - 6,997 m / s ,  3)     v = 10.15 m / s ,

4) the displacement of the second point in greater

5)     x = 0.883 m

Explanation:

For this exercise we will use the kinematics equations

1) ask which chick reaches the ground first

we calculate for the first chick that has zero initial velocity

          y = y₀ + v₀ t - ½ g t²

          0 = yo - ½ g t²

          t = √ 2 y₀ / g

let's calculate

          t = √ (2 2.50 / 9.8)

          t = 0.714 s

We calculate the time it takes for the second chick that has velocity v = (1 i ^ - 1.5 j⁾ m / s

           y = y₀ + v₀t - ½ g t²

           0 = 2.5 - 1.5 t - ½ 9.8 t²

           4.9 t² + 1.5 t - 2.5 = 0

            t² + 0.306 t - 0.510 = 0

we solve the quadratic equation

            t = [0.306 ± √ (0.306² - 4 (-0.510))] / 2

            t = [0.306 ± 1.46] / 2

The results are

            t₁ = -0.577 s

            t₂ = 0.883 m / s

we take positive time as correct

the time of the pigeon 1 is less, so it comes first

2) the speed of the first chick is

              v = v₀ - g t

we can see that

              v = -gt

              v = - 9.8 0.714

              v = - 6,997 m / s

the negative sign indicates that the speed is down

3) the speed of the other bird is

              v = -1.5 - 9.8 0.883

               v = 10.15 m / s

4) which chick has the greatest displacement. The first point falls vertically and its displacement is y₀

The second point describes a parabola and its displacement is

           d = √ (x² + y₀²)

therefore we see that the displacement of the second point in greater

5) calculate the horizontal displacement of the second point

           x = vx t

           x = 1 0.883

           x = 0.883 m

small = small current large mag large current

float.

archimedes' principle

denser than water sinks in water.

denser than mercury would sink in it