Answer:
2.75 seconds.
Step-by-step explanation:
We have been given that a ball is dropped from the ground. T represents the time in seconds.
To find the time when the ball was in air, we will equate height of ball with 0 and solve for t.
[tex]-16t^2+44t=0[/tex]
Upon dividing both sides by 4, we will get:
[tex]-4t^2+11t=0[/tex]
[tex]-t(4t-11)=0[/tex]
[tex]-t=0\text{ (or) }(4t-11)=0[/tex]
[tex]t=0\text{ (or) }4t-11=0[/tex]
[tex]t=0\text{ (or) }4t=11[/tex]
[tex]t=0\text{ (or) }t=\frac{11}{4}[/tex]
[tex]t=0\text{ (or) }t=2.75[/tex]
Since our given function is a downward opening parabola, so ball will be in air between both t-intercepts.
Since the ball touches the ground at 2.75 seconds, therefore, the ball would be in air for approximately 2.75 seconds.