107–110. {Use of Tech} Motion with gravity Consider the follow...

Question

107–110. {Use of Tech} Motion with gravity Consider the following descriptions of the vertical motion of an object subject only to the acceleration due to gravity. Begin with the acceleration equation a(t) = v' (t) = -g , where g = 9.8 m/s² .

c. Find the time when the object reaches its highest point. What is the height?

A payload is released at an elevation of 400 m from a hot-air balloon that is rising at a rate of 10 m/s.

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says a rock is thrown vertically upwards from the top of a 250 m cliff with an initial velocity of 150 m per second. Assume that only gravity acts on the rock, with G equal to 9.8 m per second squared. At what time does the rock reach its maximum height, and what is that height? So we're asked to find the time that the rock reaches the maximum height, as well as the maximum height. We were given a couple pieces of information in the problem. We're told that only gravity acts on the rock, and we're given a value for gravity, which is G is equal to 9.8 m per second squared, and this gravity is acting downward, so it's acting in the negative direction. So that means that our acceleration, AFT, which is our acceleration as time, is just going to be a constant, and it's just going to be equal to -9.8. We're also told that our initial velocity is 15 m per second, so this is our velocity at time equal to 0, so we'll have V of zero, and that's going to be equal to 15. We're also given our initial position, since we're at the top of a cliff, and that cliff has a height of 2150 m. That means that our initial position, which is going to be S of 0, is going to be equal to 250. So now we've identified the information of the problem, we can actually find the time that we take to reach the maximum height, as well as the height. So, the first thing we'll need to do is to find our velocity as a function of time, and then use that to find our position as a function of time. Now recall that our velocity is a function of time, VFT is equal to the integral of AFT. DT So here we're going to be integrating our acceleration as a function of time. However, our acceleration is just a constant, so substituting in that value for AFT. We see that VAT is going to be equal to the integral of minus 9.8 DT. And performing this integration, we see that this is going to be equal to minus 9.8 T plus D. So here we call that when you integrate a constant, you just get that constant multiplied by T, and then we still have our integration constant C. Now what we need to do is determine the value of that integration concept C, and this is where we'll use our initial conditions. So, our velocity times 0 has to be equal to 15. So that means that 15 is equal to minus 9.8, multiplied by 0, plus C. So this means that C is going to be equal to 15. And so substituting in that value for uh C, we see that VFT. is equal to -9.8 T. Plus 15. And so now we need to use our velocity as a function of time here to get our position. And we call that our position as a function of time, which is SFT. It's going to be equal to the integral of VFT. DT. So here we're integrating our velocity with respect to time. So substituting in the expression for our velocity, this is going to be equal to the interval of Minus 9.8 T. Plus 15, and then that quantity is going to be multiplied by DT. So, performing the integration for each of these terms, we see that this is going to be equal to. Minus 9.8, and here when we integrate T, we get E2d divided by 2. Then we'll have plus, and when we integrate 15 with respect to T, we get 15 T. And then we'll also have our integration cost of plus D as well. Now, again, we need to figure out the value for our integration constant, and since our initial position at T equal to 0 is 250, we will set 250. Equal to, and here we can simplify expression a little bit. So we'll have -9.8 divided by 2, which is minus 4.9, and that gets multiplied by 0 squared since T is going to be equal to 0, and we'll have plus 15 multiplied by 0, plus C. So solving this for C, we see that C is going to be equal to 250. So substituting that in, we see that S of T. It's gonna be equal to minus 4.9 E2. Plus 15 T plus 250. So here for our position equation, we have a quadratic equation with a negative coefficient in front of the T squared. So that means that our parabola is actually concave down. And so it's vertex is going to be a maximum, so we just need to find the time corresponding to the vertex of our parabola. So we just need to recall our formula for finding the vertex of a parabola. And so that means that T going to be equal to. And here we call your formula. This could be minus B divided by 2A, where B here is the coefficient of the linear term. So we'll have -15. That's gonna be divided by 2 multiplied by A, which is the coefficient of the quadratic term. So this is going to be minus 4.9. Which is going to be equal to. 15 divided by 9.8, we want to get rid of the decimal and the denominator. We'll rewrite this as being equal to. 75 divided by 49. So here we've just multiplied both the numerator and denominator by 5, and this will give us a whole number in both the numerator and denominator. So this here is the value that we're looking for for the first part of the question, since that is the time that it takes to reach the maximum height. Now, to figure out the value of that height, we just need to substitute this value for T back into our expression or our height. So, we'll have S of 75 divided by 49. And this is going to be equal to. Minus 4.9, multiplied by 75 divided by 49. Weird? We'll have plus 15 multiplied by. 75 divided by 49. Plus 250. So, evaluating this expression, this is going to be equal to. 25,625 divided by 98. Which is going to be equal to. 261. And 47 divided by 98. And that is the answer that we're actually looking for, for the second part of the problem. So just to recap, the time that it takes to reach the maximum height is 75 divided by 49 seconds. And the maximum height is 261 and 47 divided by 98 m. So those are the answers that we're looking for for this problem. So I hope that this explanation has been useful, and I'll see you in the next video.

Key Concepts

Acceleration due to Gravity

Velocity and its Relationship to Acceleration

Maximum Height in Projectile Motion

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