t Answer:
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
Step-by-step 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