Exercise 12 Page 695

We know that a rocket followed a path modeled by a quadratic equation, where t is the time in seconds. d=80t-16t^2 We are given that the rocket failed to explode, so we can determine after how many seconds the rocket would hit the ground. To do this, we will substitute 0 for d and solve the quadratic equation. 0=80t-16t^2 ⇒ -16t^2+80t=0Let's recall the Quadratic Formula. ax^2+ bx+ c=0 ⇕ x=- b± sqrt(b^2-4 a c)/2 a To solve our equation, we first need to identify the values of a, b, and c. -16t^2+80t=0 ⇕ -16t^2+ 80t+ 0=0 We see that a= -16, b= 80, and c= 0. Let's substitute these values into the Quadratic Formula.

The solutions for this equation are t= -80± 80-32. Let's separate them into the positive and negative cases.

Using the Quadratic Formula, we found that the solutions of the given equation are t_1=0 and t_2=5. Therefore, the rocket would hit the ground after 5 seconds.

Alternative Solution

Graphing Method

Let's draw the path of the rocket using the given equation. To do this, we will substitute some values for t and evaluate corresponding values of d to find the points that lie on this graph.

t	80t-16t^2	d
0	80( 0)-16( 0)^2	0
1	80( 1)-16( 1)^2	64
2	80( 2)-16( 2)^2	96
3	80( 3)-16( 3)^2	96
4	80( 4)-16( 4)^2	64
5	80( 5)-16( 5)^2	0

Now let's plot the points and connect them with a smooth curve. Remember that both time and distance are non-negative, so our graph will be drawn only in the first quadrant.

From the graph, we can see that the rocket would hit the ground after 5 seconds.