A boy stands on the edge of a cliff of height 60m. He throws a stone vertically upwards so that its distance, h, above the cliff top is given by `h = 20t - 5t^2`Calculate the time which elapses...

A boy stands on the edge of a cliff of height 60m. He throws a stone vertically upwards so that its distance, h, above the cliff top is given by `h = 20t - 5t^2`Calculate the time which elapses before the stone hits the beach.

Expert Answers
lemjay eNotes educator| Certified Educator

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.