Question

1. peregrine falcons dive on their prey with speeds up to 36 m/s. from what height must a falcon fall freely to achieve this speed?

Explanation:

Step1: Recall the free - fall formula

For an object in free - fall (starting from rest, \(v_0 = 0\ m/s\)), the final velocity \(v\) is related to the height \(h\) and the acceleration due to gravity \(g\) (where \(g= 9.8\ m/s^{2}\)) by the formula \(v^{2}=v_{0}^{2}+2gh\). Since the falcon starts from rest (\(v_0 = 0\)), the formula simplifies to \(v^{2}=2gh\). We need to solve for \(h\), so we can re - arrange the formula to \(h=\frac{v^{2}}{2g}\).

Step2: Substitute the given values

We know that \(v = 36\ m/s\) and \(g=9.8\ m/s^{2}\). Substitute these values into the formula for \(h\):
\(h=\frac{(36)^{2}}{2\times9.8}=\frac{1296}{19.6}\approx66.12\ m\)

Answer:

The falcon must fall from a height of approximately \(\boldsymbol{66.1\ m}\) (or more precisely \(\frac{1296}{19.6}\approx66.12\ m\)) to achieve a speed of \(36\ m/s\) when falling freely.