Answer:
a.2.11 s or 8.57 s
b.10.79 s
Step-by-step explanation:
We are given that
Initial velocity=171 ft/s
Height at which the object is thrown=17 ft
The height h of the object after t seconds can be found by the equation
[tex]h=-16t^2+171t+17[/tex]
a.We have to find the time taken by object to reach the height 307 ft.
Substitute the value
[tex]307=-16t^2+171t+17[/tex]
[tex]16t^2-171t-17+307=0[/tex]
[tex]16t^2-171t+290=0[/tex]
Quadratic formula :
[tex]x=\frac{-b\pm\sqrt D}{2a}[/tex]
Where [tex]D=b^2-4ac[/tex]
By using quadratic formula
[tex]t=\frac{-(-171)\pm\sqrt{(-171)^2-4(16)(290)}}{2(16)}[/tex]
[tex]t=\frac{171\pm\sqrt{10681}}{32}[/tex]
[tex]t=\frac{171\pm 103.35}{32}[/tex]
[tex]t=\frac{171+103.35}{32}=8.57[/tex]
[tex]t=\frac{171-103.35}{32}=2.11[/tex]
Hence, the object will take 8.57 s or 2.11 s to reach the height 307 ft.
b.We have to find the time taken by object to reach the ground.
When the object reach the ground, then h=0
Substitute the value
[tex]0=-16t^2+171t+17[/tex]
[tex]16t^2-171t-17=0[/tex]
Again by using quadratic formula
[tex]t=\frac{171\pm\sqrt{(-171)^2-4(16)(-17)}}{32}[/tex]
[tex]t=\frac{171\pm\sqrt{30329}}{32}=\frac{171\pm 174.15}{32}[/tex]
[tex]t=\frac{171+174.15}{32}=10.79[/tex]
[tex]t=\frac{171-174.15}{32}[/tex]
[tex]t=-0.098[/tex]
It is not possible because time cannot be negative.
Hence, the object will take 10.79 s to reach the ground.
