A toy rocket is launched with an initial velocity of 180 m/s. The height of the rocket, in metres, can be modelled by h = –5t2 + 180t, where t is the time in seconds the rocket is in the air. How long will the rocket stay above a height of 1000 m
Answer: 22.3 s
The answer says to first put in vertex form by
h = –5t2 + 180t
h=–5(t2 –36t+324–324)
h =–5(t – 18)2 + 1620
But why is it +324. And not +(36/2)^2
324 = +(36/2)^2
The expression "+324" is added in order to complete the square. Completing the square is a technique used to rewrite a quadratic expression in vertex form.
In this case, the coefficient of t in the given equation is 180. To complete the square, we divide it by 2 to get 90, then square it to get 8100. However, in order to maintain the equality of the equation, we also subtract 8100.
So, we can write the equation as:
h = -5(t^2 - 36t + 324 - 324)
Simplifying the expression inside the parentheses:
h = -5(t - 18)^2 + 1620
Now the equation is in vertex form. In this form, the vertex of the parabolic equation is represented by the values (h, k), where h is the time and k is the corresponding height.
In this case, the vertex of the parabola is (18, 1620).
To find the time the rocket will stay above a height of 1000m (h = 1000), we can substitute the value of h into the equation:
1000 = -5(t - 18)^2 + 1620
Rearranging the equation:
-5(t - 18)^2 = 1000 - 1620
-5(t - 18)^2 = -620
Dividing both sides by -5:
(t - 18)^2 = 124
Taking the square root of both sides:
t - 18 = ±√124
t - 18 = ±11.1355
Solving for t:
t = 18 + 11.1355 ≈ 29.14 seconds (ignoring the negative root)
Therefore, the rocket will stay above a height of 1000m for approximately 29.14 seconds, and not 22.3 seconds as mentioned in the initial answer.
To understand why it is +324 and not +(36/2)^2, let's break down the steps involved in converting the equation to vertex form.
We start with the equation: h = -5t^2 + 180t
Step 1: Complete the square for the t^2 and t terms.
To complete the square for the t^2 term, we need to add and subtract a quantity that allows us to write it as a perfect square. In this case, we add and subtract (180/2)^2 = 324.
h = -5(t^2 - 36t + 324 - 324) + 324
Step 2: Factor the perfect square trinomial.
Now, we group the t^2 and t terms together and factor them as a perfect square trinomial. The perfect square trinomial in this case is (t - 18)^2.
h = -5(t^2 - 36t + 324) + 324
Step 3: Simplify the equation.
At this point, we can simplify the equation by multiplying the -5 to each term within the parentheses.
h = -5(t - 18)^2 + 324
Now the equation is in vertex form. The vertex form of a quadratic equation is given by h = a(t - h)^2 + k, where the vertex of the parabola is represented by (h, k). In this case, the vertex is at (18, 324).
Therefore, after completing the square, we get h = -5(t - 18)^2 + 324. The +324 term accounts for the additional constant that was introduced when completing the square and ensures that the equation remains equivalent.