Question

a student drops a feather from a height of 26.0m. determine all unknowns and answer the following questions. how long did the feather remain in the air? what was the feathers speed just before striking the ground?

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