a student drops a feather from a height of 26.0m. determine all unknowns and answer the following questions…

Explanation:

Step1: Identify the relevant kinematic equation for time

We use the equation $h = v_0t+\frac{1}{2}gt^2$. Since the feather is dropped ($v_0 = 0$), the equation simplifies to $h=\frac{1}{2}gt^2$, where $h = 26.0m$ and $g = 9.8m/s^2$. $26.0=\frac{1}{2}\times9.8t^2$

Step2: Solve for time $t$

First, rewrite the equation as $t^2=\frac{2\times26.0}{9.8}$. Then $t=\sqrt{\frac{2\times26.0}{9.8}}\approx\sqrt{5.306}\approx2.30s$.

Step3: Identify the relevant kinematic equation for final - velocity

We use the equation $v = v_0+gt$. Since $v_0 = 0$, $v = gt$.

Step4: Calculate the final velocity $v$

Substitute $g = 9.8m/s^2$ and $t = 2.30s$ into the equation. $v=9.8\times2.30 = 22.54m/s$.

Answer:

How long did the feather remain in the air? 2.30 s What was the feather's speed just before striking the ground? 22.54 m/s