A student standing on the ground throws a ball straight up. The ball leaves the student’s hand with a speed of 15 m/s when the hand is 2.0 m above the ground. How long is the ball in the air before it hits the ground? (The student moves her hand out of the way.

The final velocity of the ball is $v_f=0$
The acceleration due to gravity is$g=10m/s^2$

Let the maximum height the ball reached is$s$

$$\begin{gathered} {v_f}^2-{v_i}^2=2gs \\ 0-{\left(15\right)}^2=2(-10)s \\ s=11.25m \end{gathered}$$

Thus, the distance travelled by the ball from girl's height till it starts coming back to ground is 11.25 m
Now, determine the time taken by ball to reach the height 11.25 m
using the equation of motion
$$\begin{gathered} v_f=v_i+gt \\ 0=15+(-10)t \\ t=1.5sec \end{gathered}$$
Thus, the ball reaches to its maximum height in 1.5 sec

It is given that the hand of girl was 2m above the ground. Therefore, while coming tot the ground, the ball travels the distance 11.25m + 2m = 13.25m
Using the equation of motion
$S=v_{i}t+\frac{1}{2}g{t}^2$
Substitute the values, $S=13.25m,v_i=0,g=10m/s^2$

role="math" $$\begin{gathered} 13.25=0+\frac{1}{2}\left(10\right)t^2 \\ {t}^2=\frac{13.25}{5}=2.65 \\ t=1.62sec \end{gathered}$$

The time taken by ball to return to ground from maximum height is 1.62 sec