# Video: Using Differentiation to Find the Maximum Height a Launched Rocket Attains

A rocket is launched in the air. Its height, in meters, as a function of time is given by ℎ(𝑡) = −4.9𝑡² + 229𝑡 + 234. Find the maximum height the rocket attains.

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### Video Transcript

A rocket is launched in the air. Its height, in meters, as a function of time is given by ℎ of 𝑡 equals negative 4.9𝑡 squared plus 229𝑡 plus 234. Find the maximum height the rocket attains.

So, we’ve been given an equation for the height ℎ of this rocket in terms of 𝑡 time. And we’re asked to determine the maximum height that the rocket attains. We can follow some key steps in order to do so. First, we need to find an expression for the first derivative of our function. In this case, that’s ℎ prime of 𝑡. By applying the power rule of differentiation, we see that ℎ prime of 𝑡 is equal to negative two multiplied by 4.9𝑡 plus 229, which simplifies to negative 9.8𝑡 plus 229.

Next, we recall that at the critical points of the function, the first derivative is equal to zero. So, we’re going to take that expression we’ve found for ℎ prime of 𝑡, set it equal to zero, and then solve the resulting equation for 𝑡. We add 9.8𝑡 to both sides and then divide by 9.8 to give 𝑡 equals 229 over 9.8, which, as a decimal, is 23.36734, or 23.37 to two decimal places. Now, this is the value of 𝑡 at which our function ℎ of 𝑡 has a critical point. We don’t yet know whether it is a maximum. And we don’t yet know the height that the rocket attains at this point.

Our next step, then, is to evaluate the function ℎ of 𝑡 when 𝑡 is equal to 23.37. Substituting into our equation ℎ of 𝑡 gives ℎ of 23.37 equals negative 4.9 multiplied by 23.37 squared plus 229 multiplied by 23.37 plus 234. Which evaluates to 2909.56119, or 2909.56 correct to two decimal places. So, we believe that this is the maximum height that the rocket attains as this occurs at the only critical point of the function. But we must confirm that it is indeed a maximum.

To do so, we’ll perform the second derivative test. We’ll evaluate the second derivative ℎ double prime of 𝑡 at this critical point. Differentiating our expression for ℎ prime of 𝑡, which was negative 9.8𝑡 plus 229, we find that ℎ double prime of 𝑡 is equal to negative 9.8. And in fact, we don’t need to evaluate the second derivative when 𝑡 equals 23.37 because it is a constant. The second derivative is the same for all values of 𝑡.

We note though, that this value is negative. And therefore, by the second derivative test, our critical point is a maximum. Now, we could also have seen this if we considered that the expression we were given for ℎ in terms of 𝑡 is a quadratic expression with a negative leading coefficient. And therefore, the graph of 𝑡 against ℎ would be an inverted parabola. And we know that it will therefore have a maximum point rather than a minimum. So, we can conclude that the maximum height the rocket attains, and we have confirmed that it is indeed a maximum, is 2909.56 meters, correct to two decimal places.