The given function is:
`h(t) = 20t-5t^2`
where h(t) represents the height of the stone above the cliff.
Since the cliff is 60m above the sea, when the stone hits the beach, the value of h(t) is -60. Plugging this value, the function becomes:
`-60 = 20t-5t^2`
Take note that to solve quadratic equation, one side should be zero.
`5t^2-20t-60=0`
The three terms have a GCF of 5. Factoring out 5, the equation becomes:
`5(t^2-4t-12)...
The given function is:
`h(t) = 20t-5t^2`
where h(t) represents the height of the stone above the cliff.
Since the cliff is 60m above the sea, when the stone hits the beach, the value of h(t) is -60. Plugging this value, the function becomes:
`-60 = 20t-5t^2`
Take note that to solve quadratic equation, one side should be zero.
`5t^2-20t-60=0`
The three terms have a GCF of 5. Factoring out 5, the equation becomes:
`5(t^2-4t-12) = 0`
Dividing both sides by 5, it simplifies to:
`t^2-4t-12=0`
Then, factor the expression at the left side of the equation.
`(t - 6)(t+ 2) = 0`
Set each factor equal to zero. And isolate the t.
`t-6 = 0`
`t=6`
`t+2=0`
`t=-2`
Since t represents the time, consider only the positive value. (Let's assume that the time t is in seconds.) So the value of t when h(t)=-60 is:
`t = 6`
Therefore, the stone hits the beach 6 seconds after it was thrown.
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