Question

a rock dropped from a cliff covers one-third of its total distance to the ground in the last second of its fall. Air resistance is negligible. How high is the cliff?

Answer 1

Hello there,

This problem caught my eye, and it was a lot of fun to puzzle my way through it. My physics one skills have gone unused for a while so this was a good way to scrape away some of the caked on rust. As challenging as this problem is at first glance, it is easy to become overwhelmed and assume that there is no what you could ever solve it. In situations like that, you need to remind yourself that problems given to you in the text book or by a teacher/prof have been designed to be solved by you, with only the skills you have been given. Hopefully that can focus you enough that you don't let nervousness get the best of you. Also, you will want to start by drawing the scenario and listing all of our givens. (first time using the draw feature, not sure if it is working)



At any rate, its a pretty simple scene, the guy on top of the cliff which has a height X, and is broken into three equal chunks; d1,d2, and d3. We know that when the rock starts at the top of the cliff, its initial velocity, V01, is zero, we know that the time it takes to pass through d3, td3, is given as 1s. We know that the acceleration is the standard 9.8m/s/s, and we know that the speed of the rock as it enters the last third of the total distance (the top of d3) is the same as the final speed as it leaves the second third of distance (the bottom of d2) - so V02 = VF1. Looking at the equations for constant acceleration we need to pick one to start with.\[1) \Delta X = V _{0} + 1/2at ^{2}\]\[2) V_{f} = V_{0} + at\]\[3) V_{f}^{2} = V_{0}^2 + 2a\Delta X\]So as we look over these equations, notice that we only have givens to fit equations 1. We know that the distance traveled was d3, we know the acceleration, and we know the time, allowing us to solve for the initial speed. Now you might say, 'Hold on...we dont know the value for d3...whats going on?' You are right, we cant obtain a number for d3, yet. But we can solve for it in terms of other variables, and then use a second step to solve for that. So our first order of business is to solve d3 in terms of the velocity at the top of d3 (which is the same as the velocity at the bottom of d2). Since the solution is complex I will show my math step by step.\[\Delta X = V_{0}t + 1/2at^2\]\[d_{3} = V_{f1}(1) + 1/2(9.8)(1)^2\]\[d_{3} = V_{f1} + 4.9\]\[d_{3} - 4.9 = V_{f1}\]Now that we have solved for Vf1, we can use equation number three from above, using the starting velocity of zero, the final velocity we just calculated, and a value of 2(d3) for our change in distance. (Remember, we solved for the velocity the rock would have at the end of passing through two thirds. This is why the distance here is two times d3) \[V_{f}^2 = V_{0}^2 + 2a \Delta x\]\[(d_{3}-4.9)^2 = 0^2 + 2(9.8)(2d_{3})\]\[d_{3}^2 - 9.8d_{3} + 24.01 = 39.2d_{3}\]\[d_{3}^2 - 49d_{3} + 24.01 = 0\]And there you have it. d3 is now in a position to be solved numerically by utilizing the quadratic formula. The quadratic formula is given as:\[(-b \pm \sqrt{b^2-4(a)(c)})/2a\]Substituting our values in for a,b, and c, the math proceeds thusly:\[d_{3}=(49\pm \sqrt{49^2-4(1)(24.01)})/2\]\[d_{3}=(49\pm \sqrt{2401-96.04})/2\]\[d_{3}=(49\pm \sqrt{2304.96})/2\]\[d_{3}=(49\pm48)/2\]\[d_{3} = 48.5m\]Now that we know what d3 is, we can simply multiply it by three to give us the total height of the cliff, 145.5m. It is always a good idea to check our work, and we can do so using equation 1) from above. Using the total height we find the travel time to be 5.4 seconds, and if we use 2/3 the height, the flight time is 4.4 seconds. This shows us that the last third of the cliff does indeed only require one second of flight time. Well done!