Question

The Vertical position y (in feet) of a rock T seconds after it was dropped from a Cliff is given by the formula y= -60tsquared +4t+380. The base of a Clift corresponds to y=0 after how many seconds with a rock hit the ground at the base of the Cliff

Answer 1

Answer:

2.48 seconds

Step-by-step explanation:

The given function is

$y=-60t^2+4t+380$

When $y=0$ the ball will be at the ground

$0=-60t^2+4t+380$

$\Rightarrow t=\dfrac{-4\pm \sqrt{4^2-4\left(-60\right)\times 380}}{2\left(-60\right)}$

$\Rightarrow t=-2.48,2.55$

So, the time taken by the rock to hit the ground is $2.48\ \text{seconds}$.