A peregrine falcon dives at a pigeon. The falcon starts downward from rest with free-fall acceleration. If the pigeon is 57.0 m below the initial position of the falcon, how long does the falcon take to reach the pigeon? Assume that the pigeon remains at rest.
h=gt²/2 =>
t =sqrt(2h/g)=
=sqrt(2•57/9.8) =3.41 s
To solve this problem, we can use the equations of motion for an object in free fall. The equation we need is:
Δy = v_i * t + (1/2) * a * t^2
Where:
Δy is the change in height (57.0 m in this case)
v_i is the initial velocity (0 m/s since the falcon starts from rest)
t is the time taken to reach the pigeon
a is the acceleration due to gravity (-9.8 m/s^2 in this case)
Now let's rearrange the equation to solve for t:
Δy = v_i * t + (1/2) * a * t^2
57.0 m = 0 * t + (1/2) * (-9.8 m/s^2) * t^2
57.0 m = (-4.9 m/s^2) * t^2
Now, we can solve for t. Going through the algebraic steps:
57.0 m = (-4.9 m/s^2) * t^2
t^2 = 57.0 m / (-4.9 m/s^2)
t^2 = -11.63 s^2
Taking the square root:
t = √(-11.63 s^2)
However, we can't take the square root of a negative number in this context since time cannot be negative. This means that the falcon cannot reach the pigeon in a straight downward free-fall. There may be other factors or forces involved that are not mentioned in the problem statement.