Answer: 4 s
Explanation:
Given
The ball leaves the hand of student with a speed of [tex]u=19\ m/s[/tex]
When the hand is [tex]h=2.5\ m[/tex] above the ground
Using the equation of motion we can write
[tex]h=ut+\dfrac{1}{2}at^2[/tex]
Substitute the values
[tex]\Rightarrow 2.5=-19t+0.5\times 9.8t^2\\\Rightarrow 4.9t^2-19t-2.5=0\\\\\Rightarrow t=\dfrac{19\pm \sqrt{(-19)^2-4\times 4.9\times (-2.5)}}{2\times 19}\\\Rightarrow t=4.0049\quad [\text{Neglecting the negative value of }t][/tex]
Thus, the ball will take 4 s to hit the ground.
