Question

A hawk flying at 13 m/s at an altitude of 156 m accidentally drops its prey. The parabolic trajectory of the falling prey is described by the equation y = 156 − x2 39 until it hits the ground, where y is its height above the ground and x is its horizontal distance traveled in meters. Calculate the distance traveled by the prey from the time it is dropped until the time it hits the ground. Express your answer correct to the nearest tenth of a meter.

Answer 1

Answer:

The distance traveled will be 156.0 m

Explanation:

Please, see the attached figure 1 for a description of the problem.

The distance traveled by the prey is the magnitude of the vector "r" (see figure).

If the components of "r" are "rx" (horizontal component) and "ry" (vertical component), its magnitude will be:

$|r| =\sqrt{(rx)^{2} + (ry)^{2}}$

We already know the vertical component of r, ry, because the prey falls from an altitude of 156 m (see figure). Then ry = 156 m

To find the value of rx we will use the function that describes the trajectory of the prey (see figure 2):

y = -39 · x² + 156

When the prey hits the ground, the height is 0 (y = 0). rx will be the value of x for which y = 0.

0 = -39 · x² + 156

-156/-39 = x²

x = 2.0 m

Then r = (2.0 m, -156.0 m)

The distance traveled will be:

$|r| =\sqrt{(2 m)^{2} + (156 m)^{2}} = 156.0 m$