Answer:
0.34 s
Explanation:
Given that,
Initial speed of a ball, u = 5.8 m/s
It is kicked at an angle of 17.0° above the horizontal.
The vertical component of velocity will be,
[tex]u_y=u\sin\theta\\\\=5.8\times \sin 17\\\\=1.69\ m/s[/tex]
Let it takes t time in the air before it lands on the ground again. It can calculated as :
[tex]t=\dfrac{2u_y}{g}\\\\t=\dfrac{2\times 1.69}{9.8}\\\\t=0.34\ s[/tex]
So, it will take 0.34 seconds.
