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 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 $h = 21.0m$ and $g = 9.8m/s^2$.
$h=\frac{1}{2}gt^2\Rightarrow t=\sqrt{\frac{2h}{g}}$

Step2: Calculate the time

Substitute $h = 21.0m$ and $g = 9.8m/s^2$ into the formula.
$t=\sqrt{\frac{2\times21.0}{9.8}}\approx\sqrt{4.286}\approx2.07s$

Step3: Identify the relevant kinematic - equation for final velocity

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

Step4: Calculate the final velocity

Substitute $t = 2.07s$ and $g = 9.8m/s^2$ into the formula.
$v=9.8\times2.07 = 20.29m/s$

Answer:

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