Question

a student drops a feather from a height of 21.0m. 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 $h = v_0t+\frac{1}{2}gt^2$. Since the feather is dropped, $v_0 = 0$. The equation simplifies to $h=\frac{1}{2}gt^2$, where $h = 21.0m$ and $g = 9.8m/s^2$.
$$21.0=\frac{1}{2}\times9.8\times t^2$$

Step2: Solve for time $t$

First, rewrite the equation as $t^2=\frac{2\times21.0}{9.8}$. Then $t=\sqrt{\frac{2\times21.0}{9.8}}\approx2.07s$.

Step3: Identify the relevant kinematic equation for final - speed

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

Step4: Calculate the final speed $v$

Substitute $g = 9.8m/s^2$ and $t\approx2.07s$ into $v = gt$. So $v=9.8\times2.07 = 20.29m/s$.

Answer:

Time in air: $2.07s$
Speed just before striking the ground: $20.29m/s$