Answer:
t1= 5.35
Step-by-step explanation:
Projectile height formula should be
h= -1/2gt^2+v0t+h0
where
h= height, this question asking time when "ball hit the ground" so height is 0
t= time, we look for this variable
v0= initial velocity= 85 feet/s
h0= initial height = 6 feet
g= gravity, assumed 32 feet/s^2
The calculation will be:
h= -1/2gt^2+v0t+h0
0= -[tex]\frac{1}{2}[/tex] 32 t^2+85t+6
0= -16t^2+85t+6
16t^2 - 85t -6 =0
x= [tex]\frac{-b±\sqrt{ b^2-4ac}}{2a}[/tex]
t= [tex]\frac{(-(-85)±\sqrt{ -85^2-4(16)(-6)}}{2(16)}[/tex]
t= [tex]\frac{(-(-85)±\sqrt{7417}}{2(16)}[/tex]
t= [tex]\frac{85±\sqrt{7417}}{32)}[/tex]
t= [tex]\frac{85±\ 86.12}{32}=[/tex]
t1= [tex]\frac{85+\ 86.12}{32}=[/tex]
t1= 5.35
t2= [tex]\frac{85-\ 86.12}{32}=[/tex]
t2= -0.035
The answer can't be minus, so its t1
