Answer:1.32 s
Step-by-step explanation:
Given
The initial height of ball H=2 m
Initial velocity u=5 m/s
Height of ball at any instant is given by
[tex]H=-4.9t^2+5t+2[/tex]
The ball will hit the ground when h=0
[tex]-4.9t^2+5t+2=0\\4.9t^2-5t-2=0\\t=\dfrac{5\pm\sqrt{25+4(4.9)(2)}}{2\times 4.9}\\t=\dfrac{5\pm\sqrt{64.2}}{9.8}\\t=\dfrac{5+8.01}{9.8}=1.32\ s\quad\text{Neglecting negative value of t}[/tex]