Answer:
[tex]t=2.47\ s[/tex]
The ball takes 2.47 seconds to touch the ground
Step-by-step explanation:
The equation that models the height of the ball in feet as a function of time is:
[tex]h(t) = h_0 + s_0t -16t ^ 2[/tex]
Where [tex]h_0[/tex] is the initial height, [tex]s_0[/tex] is the initial velocity and t is the time in seconds.
We know that the initial height is:
[tex]h_0 = 4.5\ ft[/tex]
The initial speed is:
[tex]s_0 = 110sin(20\°)\\\\s_0 = 37.62\ ft/s[/tex]
So the equation is:
[tex]h (t) = 4.5 + 37.62t -16t ^ 2[/tex]
The ball hits the ground when when [tex]h(t) = 0[/tex]
So
[tex]4.5 + 37.62t -16t ^ 2 = 0[/tex]
We use the quadratic formula to solve the equation for t
For a quadratic equation of the form
[tex]at^2 +bt + c[/tex]
The quadratic formula is:
[tex]t=\frac{-b\±\sqrt{b^2 -4ac}}{2a}[/tex]
In this case
[tex]a= -16\\\\b=37.62\\\\c=4.5[/tex]
Therefore
[tex]t=\frac{-37.62\±\sqrt{(37.62)^2 -4(-16)(4.5)}}{2(-16)}[/tex]
[tex]t_1=-0.114\ s\\\\t_2=2.47\ s[/tex]
We take the positive solution.
Finally the ball takes 2.47 seconds to touch the ground
