(II) A helicopter is ascending vertically with a constant spee...

Question

(II) A helicopter is ascending vertically with a constant speed of 6.40 m/s. At a height of 105 m above the Earth, a package is dropped from the helicopter. How much time does it take for the package to reach the ground? [Hint: What is v₀ for the package?]

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem as a hot air balloon rises vertically at a steady rate of 8.0 m per second. A sandbag is dropped from the balloon when it reaches an elevation of 300 m above the ground, determine the time it takes for the sandbag to descend to the ground. So that's her end goals. We're trying to figure out how much time it's gonna take for the sandbag to fall from the hot air balloon to the ground. OK. And we're also given some multiple choice answers. They're all in the same units of seconds. So let's read them off to see what our final answer will be. A is 8.7 B is 7.1 C is 5.9 and D is 5.1. OK. So first off, let us consider the upwards motion to be positive. Now, we need to recall and use the Kams equation to solve for the time required for the sandbag to fall to the ground. And that states that the change in distance which is delta Y is equal to the initial velocity multiplied by the time plus one half multiplied by the acceleration multiplied by the time squared. So off to the side here it was write it in blue. Let us note that delta, why the change in distance is equal to the final distance minus the initial distance which is equal to. So the final distance is 0 m minus our initial distance, which is 300 m. So that would mean that our delta Y is equal to negative 300 meters. And let's also note that our initial velocity is equal to 8.0 m per second. And let's also note that our acceleration in this case is due to gravity and gravity is trying to pull the sandbag to the gravis gravity is pushing down on all objects. So it's gonna be negative 9.81 m per second squared. Awesome. So now we can plug in all of our known variables and solve for the amount of time it will take for the sandbag to fall to the ground. So let's do that. So we're gonna have to plug in all of our known variables and then we're gonna have to rearrange our equation to solve for T. So let's do that. So our delta Y is negative 300 m is equal to our initial velocity which is 8.0 m per second, multiplied by T the time plus one half multiplied by negative 9.81 m per second squared, multiplied by T squared. OK. Now, we need to rearrange, like I said to solve for T. So when we do our rearranging, we will get using a little bit of algebra, we'll get that 4.905 T squared. So 4.905 T squared minus 8.0 meters per second minus 300 m is equal to zero. And this should look very familiar because this is a quadratic function. So we need to apply the quadratic formula to this, to this equation, this mathematical expression. And we will get that T is equal to like negative parenthesis, negative 8.0 meters per second plus or minus the square root of negative 8.0. But don't forget our units, 8.0 m per second squared minus four, multiplied by 4.905 multiplied by negative 300 meters all divided by two, multiplied by 4.905. OK. Awesome. So we cannot have a negative value for our seconds. We can't have a negative second. So our final and correct answer must be T is equal to 8.6785 seconds, which we can round to one decimal place. And when we do that, we will get that T is equal to 8.7 seconds because as we can see in our multiple choice answers, all of our answers are to one decimal place. So that's why we needed to round to one decimal place. Hooray, we did it, we found our final answer. So looking at our multiple choice answers, the correct answer has to be the letter A which is 8.7 seconds. Thank you so much for watching. Hopefully that helped and I can't wait to see you with the next, next problem. Thanks for watching. Bye.

(II) A helicopter is ascending vertically with a constant speed of 6.40 m/s. At a height of 105 m above the Earth, a package is dropped from the helicopter. How much time does it take for the package to reach the ground? [Hint: What is v₀ for the package?]

Key Concepts

Free Fall

Initial Velocity

Kinematic Equations

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