Answer:
19.0 s
Explanation:
Since the motion of the ball is a free-fall motion, we can solve the problem by using the following suvat equation:
[tex]s=ut+\frac{1}{2}at^2[/tex]
where, chosing upward as positive direction, we have:
s = -250 m is the displacement of the ball to reach the ground
u = +80 m/s is the initial velocity of the ball
t is the time
[tex]a=g=-9.8 m/s^2[/tex] is the acceleration of gravity
Re-writing the formula we have:
[tex]-250 = 80t -\frac{1}{2}(9.8)t^2\\4.9t^2-80t-250=0[/tex]
This is a second-order equation, so we solve it by applying the usual formula to find t:
[tex]t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-(-80)\pm \sqrt{(-80)^2-4(4.9)(-250)}}{2(4.9)}[/tex]
And considering only the positive solution, from the calculation we get
t = 19.0 s
So, the ball will be in air for 19.0 s.
