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 .

d. Find the time when the object strikes the ground.

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 that a rock is thrown vertically upwards from the top of a 250 m cliff with an initial velocity of 15 m per second. Assume only gravity acts on a rock, with G equal to 9.8 m per second squared. When does the rock hit the ground at the base of the cliff? First thing we want to do is identify the information that we're given to the problem. So one piece of information that we're given is to assume that only gravity acts on the rock, and we're given that the strength of gravity G is equal to 9.8 m per second squared. Now recall that gravity is pulling everything downwards, so that's going to be the negative direction. And gravity is an acceleration, so that means that our acceleration as a function of time, AFT it's just gonna be equal to a constant, and that's going to be equal to -9.8. Now we're also told our initial velocity, and the problem, and that is 15 m per second. That means that our initial velocity be of 0, since it's the velocity at time equal to 0, and that's just gonna be equal to 15. We're also given our initial position. Since we're throwing the rock at the top of the cliff, and that's 250 m above the ground, and so our initial position, S of 0, is going to be equal to 250. So, in order to find the time at which the rock hits the ground at the base of the cliff, we first need to find our position as a function of time. In order to do that, we'll need our velocity as a function of time. So recall that our velocity is a function of time, VFT. That is going to be equal to the integral of AFT. DT. So here we're integrating our acceleration with respect to time. Now our acceleration is constant, so we'll just substitute in that value into this formula. So this is going to be equal to the integral of minus 9.8 dT. So here we're integrating a constant. I recall that when you get integrated constant, you get that constant multiplied by T. So this is going to be equal to minus 9.8 multiplied by T. Then we'll have to have our integration constant as well. So we'll have plus C. So the first thing we want to do is to figure out the value of this integration constant, and for this, we'll use our initial conditions. So we're told when at T equal to 0, our velocity should be equal to 15. So sub CT and 15 for VFT we'll have 15 equal to minus 9.8, multiplied by 0, since we're setting T equal to 0, then we'll have plus C. And so, solving this for C, we see that C is equal to 15. So, substituting that value back in for C, we have that VFT. is equal to minus 9.8 T plus 15. So now what we're going to do is use this velocity as a function of time to figure out our position. And so we call that our position as a function of time, which is going to be SFT, that that is going to be equal to the integral of our velocity, BFT. With respect to tea. So substituting in the expression that we've just found for VFT, this is going to be equal to the integral of the quantity of minus 9.8 T plus 15 in quantity DT. So integrating each of these terms one by one. This integral is going to be equal to. Minus 9.8, multiplied by, and here we integrate T with respect to T and that will give us T squared, divided by 2. Then we'll have plus, and here we're integrating 15 with respect to T and that'll just give us 15T. And then we'll have plus C since we still need an integration constant. And again, we need to figure out the value of this constant. And just like before, we're going to use our initial condition. So, at T equal to 0, our initial height is 250. So we'll have 250 is equal to, and here we can simplify the first term in our expression. 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, then we'll have plus 15 multiplied by 0, plus C. So, solving this for C, we see that C is equal to 250. That means that our expression for SFT. It's going to be equal to minus 4.9 T2. Plus 15. Plus 250. So now we use this position equation equation to find the time at which the rock hits the ground. Now, the rock is going to hit the ground when S of T is equal to 0. So setting S of T equal to 0 and solving for T will have 0 is equal to minus 4.9 T squad. Plus 15 T. Plus 250. So here we have a quadratic equation. And so we can actually use the quadratic formula in order to find T. So, we call you formula for the quadratic equation, we're gonna have that T is equal to. Minus 15, plus or minus the square root of the quantity of 15 squared. -4, multiplied by -4.9. Multiplied by 250. In quantity, and all of that is divided by the quantity of 2 multiplied by -4.9. So simplify this expression, we get that it is equal to minus 15, plus or minus the square root. Of 5,125, that quantity is going to be divided by minus 9.8. So this means we'll get two values for T. We'll get a T. is equal to Minus 15 plus the square root of 5,125. That quantity is going to be divided by minus 9.8, and we'll get that T is equal to the quantity of minus 15 minus the square root of 5,125 in quantity, divided by minus 9.8. Now, the first value of T, one with the plus sign, is actually going to be a negative quantity. Since the square root of 5,125 is greater than 15. So that means our numerator will be positive, but the denominator is negative, so that will give us an overall negative sign. And a negative time is not physically relevant, since that would correspond to an action or a time before we actually threw the rock. And so the rock can't hit the ground before we throw it. So that means that our second time there, the one with the minus sign, is going to give us the positive time. So, we'll just evaluate that second time, since that is the only physically relevant answer. And so we'll just need to simplify this expression. So we'll have T going to be equal to, and here we want to get rid of the decimal on the denominator. We'll multiply both the numerator and denominator by 10. We'll have 10 multiplied by the quantity of minus 15, minus 5 multiplied by the square root of 205 In quantity, that's gonna be divided by 98. So we can factor out a -5 out of the numerator, and so the minus signs will cancel. This is going to be equal to. 50 Divided by 98. Multiplied by the quantity of 3 plus the square root of 205. In quantity and simplifying this fraction in front by dividing both the numerator and denominator 2, this is going to be equal to 25 divided by 49, multiplied by the quantity of 3 plus the square root of 205. In quantity. So that is the answer that we're looking for for this problem, since our other time value actually gave us a negative time value, which was not relevant for this problem. So this was the only answer that we're looking for. So, I hope that this explanation has been useful, and I'll see you in the next video.

Key Concepts

Acceleration due to Gravity

Kinematic Equations

Initial Conditions

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