A rock is tossed straight up from ground level with a speed of 20 m/s. When it returns, it falls into a hole 10 m deep.
a. What is the rock’s velocity as it hits the bottom of the hole?
b. How long is the rock in the air, from the instant it is released until it hits the bottom of the hole?

The rock is tossed straight up to the ground with the initial velocity $v_i=20m/s$
The final velocity before it starts returning to the ground,$v_f=0$$v_f=0$
The acceleration due to gravity is $g=10m/s^2$

Using the equation of motion

$$\begin{gathered} {v_f}^2-{v_i}^2=2aS \\ 0-{\left(20\right)}^2=2(-10)S \\ S=\frac{400}{-20}=20m \end{gathered}$$
The distance traveled by the rock before it starts returning back to the ground is 20m

The time taken to travel this distance,

$$\begin{gathered} v_f=v_i+at \\ 0=20-10t \\ t=2sec \end{gathered}$$

Thus, the time taken by a rock to reach maximum height is 2 sec.

Now, to determine the rock's velocity before hitting the ground, analyze the distance traveled by the rock. Since it falls in the deep hole of 10 m, therefore, the total distance traveled by rock once it starts returning towards the ground is 20m +10m = 30m

By using the equation of motion,
$$\begin{gathered} {v_f}^2-{v_i}^2=2as \\ {v_f}^2-0=2\left(10\right)\left(30\right) \\ {v}_f=\sqrt{600}=24.5m/sec \end{gathered}$$

Firstly, determine the time taken by the rock to hit the bottom of the hole from the point it starts returning back into the air.
By using the equation of motion,

$v_f=v_i+g{t}_2$
The initial velocity of rock at that point is 0 m/sec while the final velocity is determined in the previous step i.e. $v_f=24.5m/sec$
Substitute the values in the above expression,

$$\begin{gathered} 24.5=0+10t_2 \\ {t}_2=2.45sec \end{gathered}$$

The total time period the rock has taken to stay in the air;

$$\begin{gathered} t=t_1+t_2 \\ t=2+2.45=4.45sec \end{gathered}$$

Therefore, the total time taken by the rock is 4.45 sec.