{Use of Tech} Launching a rocket A small rock...

Question

{Use of Tech} Launching a rocket A small rocket is launched vertically upward from the edge of a cliff $80$ ft above the ground at a speed of $96$ ft/s. Its height (in feet) above the ground is given by $h$\left$(t$\right$)=-16t^2+96t+80$ , where $t$ represents time measured in seconds.
a. Assuming the rocket is launched at $t=0$ , what is an appropriate domain for $h$ ?

a. Assuming the rocket is launched at $t=0$ , what is an appropriate domain for $h$ ?

Accepted Answer

Welcome back everyone. In this problem, from the edge of a sea cliff 25 meters above sea level, a missile is launched vertically upward at a speed of 36 meters per second. The function h of t equals negative 9 t squared plus 36 t plus 25 represents its height above sea level, where t is time in seconds. And we want to provide the domain of h if the missile is launched at t equals 0. Now first, let's try and make sense of this problem. What's what's really going on here? Well, we have a missile that's been launched from a cliff that's 25 meters above sea level. So I'm gonna draw my little cliff here and I'm gonna draw some waves. Right? Here's my little sea waves. And what do we know? We know that our missile is launched and models the function h of t equals negative 9 t squared plus 36 t plus 25 where h of t is its height above sea level and the t is the time in seconds. So in other words, if we were to look at the trajectory of our missile, which would form a parabola, then it can be modeled by the function h of t while it is in the air. And we also know that the missile is launched at t equals 0. In our problem, we want to figure out the domain of h and what that means is that we want to find the largest and smallest possible input values for our function h. In other words, we want to find the largest and smallest values of t. Now, we already know that the missile was launched at t equals 0, so that's the smallest possible value. But, what would be the highest possible value for t? Well, if we think about launching the missile, eventually, it's going to have to hit the sea surface And when it hits the sea surface, then the height of our missile is going to be equal to 0. So if we call t s the time the missile hits the sea surface, then at that time t s, the height will be equal to 0. Therefore, when the height equals 0, that would be the highest possible value of t. The missile hits the c. So if we can figure out the time at that point, we'll be able to figure out the highest value in our domain. So let's think about it. We already have the function, so we can substitute that value to solve for t. So substituting h of t equals 0 to solve for the time the missile hits the sea surface, then our function will now become 0 equals negative 9 t squared plus 36 t plus 25. Now we can solve for t. Notice that our function is a quadratic function and it's already written in the general form. So we can solve by using the quadratic formula because recall that the quadratic formula says that if we have a quadratic function in its general form, then our unknown value x equals negative b plus or minus the square root of b squared minus 4ac all divided by 2a. From our equation here, we note that a, the coefficient of the squared term equals negative 9. B, the coefficient of the linear term equals 36. And c, the constant equals 25. So substituting it into our quadratic formula, then our unknown value t would be equal to negativebnegative36 plus or minus the square root of 36 squared minus 4 multiplied by negative 9 multiplied by 25. And all of that is being divided by 2 multiplied by a, which is negative 9. So let's go ahead and solve for t. Now, negative 36 plus or minus the square root of 36 squared. 36 squared is well, let's leave the 36 squared and you're going to see why in a minute. Okay? So we have 36 squared. Negative 4 multiplied by negative 9 is 36. So we have 36 multiplied by 25, and all of that is being divided by negative 18. Now, 36 is a factor of both terms under the square root. So, we can rewrite it as negative 36 +inus36multipliedby36 +25 all divided by negative 18 and 36 +25 equals 61. So, all of that is still being divided by negative 18. Now we can take it a step further. Let me shift to the right here to continue. So that means t equals negative 36 plus or minus the square root of 36 multiplied by the square root of 61 all divided by negative 18. And, the square root of 36 equals 6. So, I can write that as negative 36 plus or minus 6 multiplied by the square root of 61 all divided by negative 18. Now, we can go ahead and divide through by negative 18 and negative 36 divided by negative 18 is 2. 6 divided by negative 18 is negative 1 third. So we have 2 plus or minus negative 1 thirds of the square root of 61 which is a negative square root of 61 divided by 3. Now, let's think about it. Okay? So, this is our value for t. There are 2 possible values. This tells us that either t equals 2 plus negative square root of 61 divided by 3, which should have been 2 minus the square root of 61 divided by 3 or t equals 2 minus negative square root of 61 divided by 3 which becomes 2 plus the square root of 61 divided by 3. By using your calculator, 2 minus the square root of 61 divided by 3 equals negative 0.6 and 2 plus the square root of 61 divided by 3 is approximately equal to 4.6 and both of them are approximations. Now, let's think about it. Now, we have some values for t. But the problem is t cannot be a negative value. We can't have a negative value for time. So, therefore, t would have to be equal to approximately 4.6 seconds or 2 plus the square root of 61 divided by 3. So that would be the upper limit for our domain. And that tells us, therefore, that the domain for h so, therefore, the domain for h is where we started 0 seconds and where the missile hits the sea surface exactly 2 plus the square root of 61 divided by 3 seconds and that would be the domain of h. The set of all non negative real numbers that represent the time from when the missile is launched until it hits the sea surface. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Quadratic Functions

Domain of a Function

Projectile Motion

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