Answer:
Option A is correct
Step-by-step explanation:
The given equation is
h = -16t^2+6t+4
When the ball will hit the ground, height h = 0
Putting value of h = 0
0 = -16t^2+6t+4
Now solving to find the value of t
=> -16t^2+6t+4 = 0
Multiply with -1
16t^2-6t-4=0
Using quadratic formula to find value of t
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
a= 16, b= -6 and c= -4
Putting values,
[tex]t=\frac{-(-6)\pm\sqrt{(-6)^2-4(16)(-4)}}{2(16)}\\t=\frac{6\pm\sqrt{36+256}}{32}\\t=\frac{6\pm\sqrt{292}}{32}\\t=\frac{6\pm17.08}{32}\\t=\frac{6+17.08}{32} \,\,and\,\, t=\frac{6-17.08}{32}\\t=0.721\,\,and\,\, t=-0.34[/tex]
Since time cannot be negative, so t = 0.721 or 0.72
So, Option A is correct