Answer:
Time when the height is 4 feet is given by
[tex]t= 0.82[/tex] sec or [tex]t=0.30[/tex]sec
and time taken to reach the ground is 1.125 sec
Step-by-step explanation:
It is given that the height is given by
[tex]h= -16t^2 +18t[/tex]
now we need to find the time t when height h= 4 feet , so we plug h= 4 and solve for t, so we have
[tex]4=-16t^2 +18t[/tex]
[tex]16t^2-18t+4=0[/tex] (adding [tex]16t^2[/tex] and subtracting [tex]18t[/tex] to both sides)
now we can use quadratic formula
[tex]t=\frac{-b\pm\sqrt{b^2-4ac} }{2a}[/tex]
we compare [tex]16t^2-18t+4=0[/tex] with [tex]at^2+bt+c=0[/tex]
we have [tex]a=16 , b=-18, c=4[/tex]
Now we have
[tex]t=\frac{-(-18)\pm\sqrt{(-18)^2-4(16)(4)} }{2(16)}[/tex]
[tex]t=\frac{18\pm\sqrt{68} }{32}[/tex]
[tex]t=\frac{18\pm8.246 }{32}[/tex]
[tex]t=\frac{18+8.246}{32}[/tex] or [tex]t=\frac{18-8.246}{32}[/tex]
so the time when the height is 4 feet is given by
[tex]t= 0.82[/tex] sec or [tex]t=0.30[/tex]sec
Now height is 0 on the ground, so we plug h=0 to find t
[tex]0=-16t^2 +18t[/tex]
[tex]16t^2 -18t=0[/tex]( we bring all the terms to left side)
[tex]2t(8t-9)=0[/tex]
[tex]t=0 [/tex] or [tex]8t-9=0[/tex] ( we equate each factor to 0)
[tex]t=0 [/tex] or [tex]t=\frac{9}{8} = 1.125[/tex]
t=0 is the initial time
hence time taken to reach the ground is 1.125 sec
