I'm not sure how to solve this.

Question

A rocket is launched from the top of a cliff that is 80 feet high with an initial velocity of 240
feet per second. The height, h(t), of the rocket after t seconds is given by the equation
h(t) = -16t^2 + 240t + 80. How long after the rocket is launched will it strike the ground?
Round to the nearest tenth of a second, if necessary.

Answer 1

You can solve this by doing the quadratic formula.  If you use the quadratic formula you will get two solutions (one negative and one positive).  In the context you can't have time to be negative, so it will be the positive answer.

 

You can solve this by graphing.  The solution will be where the graph crosses the x-axis on the positive side because time can't be negative.

 

Either way you will t =15.326 seconds

Answer 2

The rocket will strike the ground when h(t) = 0

 

-16t² + 240t + 80 = 0

-16(t² - 15t - 5) = 0

t² - 15t - 5 = 0

 

t = 15 ± √[15² - 4(1)(-5)]

                 2(1)

 

t = 15 ± √(225 + 20)

               2

 

t = 15 ± √245

           2

 

Because t is a unit of time, we must choose the root t = (15 + √245)/2

 

√245 = 15.7

 

t = (15 + 15.7)/2 = 30.7/2 = 15.4 seconds