Answer:
4.643 seconds
Step-by-step explanation:
We have been given that
s = 6 feet
v = 73 feet per second
Substituting these values in the formula [tex]H(t)=-16t^2+vt+s[/tex]
[tex]H(t)=-16t^2+73t+6[/tex]
When the ball hits the ground, the height becomes zero. Thus, H(t)=0
[tex]-16t^2+73t+6=0[/tex]
We solve the equation using quadratic formula [tex]x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
Substituting the values [tex]a=-16,\:b=73,\:c=6[/tex]
[tex]t_{1,\:2}=\frac{-73\pm \sqrt{73^2-4\left(-16\right)6}}{2\left(-16\right)}\\\\=\frac{-73\pm\sqrt{5713}}{-2\cdot \:16}\\\\=-0.081,4.643[/tex]
time can't be negative. Hence, t = 4.643.
Hence, the ball will take 4.643 seconds to hit the ground.
