Comparing velocities Two stones are thrown vertically upward, ...

Question

Comparing velocities Two stones are thrown vertically upward, each with an initial velocity of 48 ft/s at time t=0. One stone is thrown from the edge of a bridge that is 32 feet above the ground, and the other stone is thrown from ground level. The height above the ground of the stone thrown from the bridge after t seconds is f(t) = − 16t²+48t+32. and the height of the stone thrown from the ground after t seconds is g(t) = −16t²+48t.
a. Show that the stones reach their high points at the same time.

a. Show that the stones reach their high points at the same time.

Accepted Answer

Hi, everyone. Let's take a look at this practice problem. This problem says 2 rockets are launched vertically with the same initial speed of 60 m per second. Rocket A is launched from a platform 45 m high, and rocket B is launched from the ground. The height of rocket A after T seconds is given by FFT, which is equal to -5 T2 + 60T + 45. And the height of rocket B is given by G of T, which is equal to minus 5T 2 plus 60T. At what time do both rockets reach their maximum height? We're getting 4 possible choices as our answers. For choice A, we have 18 seconds. for choice B, we have 16 seconds. For C, we have 8 seconds. And for choice D, we have 6 seconds. Now we need to find the time at which both rockets reach their maximum height. So to find the time at which each rocket reaches its maximum height, we will need to first find the velocity as a function of time for that rocket, and then set that expression equal to zero and solve for a time, since our rocket will reach the maximum height when our velocity is equal to 0. So, we're gonna start off by looking at Rocket A. So we need to first find its velocity as a function of time, so we'll call that Velocity VA. And that's gonna be a function of time, and that's gonna be equal to the derivative with respect to T of our position equation, which is F of T. So this is going to be equal to the derivative with respect to T of the quantity of minus 5 T squared, plus 60T plus 45. And so taking this derivative, this is going to be equal to. For the first term, we're going to use our constant multiple and power rule. And so when we take its derivative, we will have -5 multiplied by 2, which is gonna be -10. Let's gets multiplied by T raised to the 2 minus 1, which is T to the 1, which is just T. Then we'll have plus, the derivative with respect to T of 60T. When I take that derivative, I will just get 60. And then we'll have the derivative of 45 with respect to T. So here I'm taking the derivative of constant, so it's going to be equal to 0, so we'll have plus 0. That means that this is going to be equal to minus 10 T plus 60. So now that we have our velocity as a function of time for rocket A, we'll set it equal to 0. So, we'll have -10 T. Plus 60 is equal to 0, and now we just need to solve for T. So we can add 10 T to both sides, and so we'll have 60 is equal to 10 t and divide both sides by 10, we'll get T is equal to 6 seconds. Our time here is in seconds since we're told that T is given in seconds. Now, we need to repeat this process for rocket B. So, we'll need to find our velocity of rocket B, which we'll label as VB as a function of time. And this is going to be equal to the derivative with respect to T. Of the function GFT, since that is the position function for rocket B. So this is going to be equal to the derivative with respect to T. Of the quantity of minus 5 T squared plus 60T. And so this is going to be equal to, when I take a derivative of that first term, I'm gonna get minus 10T. And when I take the derivative of the second term, I will get plus 60. So now I need to set this equation equal to 0. So we'll have minus 10 T plus 60 is equal to 0, and this is the same expression that we had for rocket A. And so, since we've already solved it, we know that T is going to be equal to 6 seconds. So, in both cases, the time to reach the maximum height for each rocket is 6 seconds. So, now if I look at my answer choices, we see that the correct answer choice corresponds to answer choice D. So I hope that this explanation has been useful, and I'll see you in the next video.

Key Concepts

Quadratic Functions

Vertex of a Parabola

Initial Velocity and Gravity

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