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

2.48 seconds

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=(-4\pm √(4^2-4\left(-60\right)* 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}`.