A projectile is fired horizontally with speed v_o from the top of the cliff of h

Question

A projectile is fired horizontally with speed v_o from the top of the cliff of height h. It immediately enters a fixed tube with length x. There is a friction between the projectile and the tube, the effect of which is to make the projectile decelerate with a constant acceleration -a (a is a positive quantity). After the projectile leaves the tube, it undergoes normal projectile motion down to the ground What is the total horizontal distance l in the figure above, that the projectile travels, measured from the base of the cliff? Give your answer in terms of x, h, v_0, g, and a. Note: This problem is simpler than it seems; make sure you don't overthink the problem. Suppose you arc given the following values: x = 0.5 m. h = 1.2 m. v_0 = 10 m/s, g = 9.81 m/s^2, and a = 2.0 m/s^2. What is the distance from the base of the cliff to the place where the ball hit the ground? Suppose you are given the following values: x = 0.5 m, h = 1.2 m, v_0 = 10 m/s, g = 9.81 m/s^2, and a = 3.0 m/s^2. What is the distance from the base of the cliff to the place where the ball hit the ground? Notice the only thing that changed between parts (b) and (c) is the magnitude of the deceleration associated with the tube. Do the differences in the answers from parts (b) and (c) makes sense with the change in a? i.e. using the results from parts (b) and (c) does your equation from part (a) make physical sense if the only value that is changed is a? Explain your answer in detail. Find the value of x and >>maximizes the equation you found in part (a). (a) with respect to x. Set the derivative equal to zero and solve for x. Remember to check if it is a maximum or minimum by graphing or testing the x value, or by taking another derivative. It may be helpful to know: d/dx (square squareroot of m + nx) = 1/2 n/square squareroot of m + nx

Explanation / Answer

5. Looking at the two terms in above equation we see that we have competing effects of x. Increasing x increases by having the projectile motion start farther to the right. But increasing x also decreases the projectile motion’s initial speed and hence its horizontal distance. Maximizing the in equation by taking the derivative with respect to x gives

0 = d/dx = 1 + ½( (2a/ (v0^2 – 2ax) ) (2h/ g) )= (v0^2 2ax) = a (2h/g) = x = v0^2/ 2a – ah/g . Both terms here correctly have dimensions of length. This result for the optimal value of x is smaller than v0^2 /2a, as it should be, because otherwise the projectile would reach zero speed inside the tube (since v = (v0^2 2ax)) and never make it out. However, we aren’t quite done, because there are two cases to consider. To see why, note that if a = v0 ( g/2h), then the x equals zero. So this equation is applicable only if a v0 g/2h. If a v0 g/2h, then it yields a negative value of x, which isn’t physical. The optimal x in the a v0 g/2h case is therefore simply x = 0. Physically, if a is large then any nonzero value of x will hurt (by slowing down the initial projectile speed) more than it will help (by adding on the “head start” distance of x). Mathematically, in this case the extremum of occurs at the boundary of the allowed values of x (namely, x = 0), as opposed to at a local maximum; a zero derivative therefore isn’t relevant.

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