Explanation:
It is given that,
Speed of the ball, u = 5 m/s
When the ball reaches the same height as that of its initial height, the vertical displacement is equal to zero i.e. [tex]s_y=0[/tex]
Let t is the time for which the ball stay in the air until it caught again at the same height. Using second equation of motion as :
[tex]s_y=ut+\dfrac{1}{2}at^2[/tex]
Here, a = -g
[tex]ut-\dfrac{1}{2}gt^2=0[/tex]
[tex]5t-\dfrac{1}{2}\times 9.8t^2=0[/tex]
On solving the above quadratic equation, we get the value of t as :
t = 1.02 seconds
So, the ball stay in the air for 1.02 seconds. Hence, this is the required solution.
