Answer:
2.48 seconds
Step-by-step explanation:
The given function is
[tex]y=-60t^2+4t+380[/tex]
When [tex]y=0[/tex] the ball will be at the ground
[tex]0=-60t^2+4t+380[/tex]
[tex]\Rightarrow t=\dfrac{-4\pm \sqrt{4^2-4\left(-60\right)\times 380}}{2\left(-60\right)}[/tex]
[tex]\Rightarrow t=-2.48,2.55[/tex]
So, the time taken by the rock to hit the ground is [tex]2.48\ \text{seconds}[/tex].
