a student drops a feather from a height of 21.0m. determine all unknowns and answer the following questions…

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: $20.29m/s$