A sandal is dropped from the top of a 15.0-m-high mast on a ship moving at 1.75 m/s due south. Calculate the velocity of the sandal when it hits the deck of the ship: (a) relative to the ship and (b) relative to a stationary observer on shore. (c) Discuss how the answers give a consistent result for the position at which the sandal hits the deck.
c) The position where the sandal hits the deck of the ship depends on how the x-component of the velocity of the sandal compares with the x-component of the velocity of the ship. From the perspective of a person on the ship, neither the sandal nor the ship have any velocity in the x-direction, and from their perspective the sandal lands straight down from where it started. From the perspective of someone on the shore, the sandal and the ship have the exact same x-component of each of their respective velocities. They each travel the same distance horizontally after the sandal is dropped, and so the sandal lands on the deck at the position directly below where it started -- the same answer in other words compared with the observer on the ship.
OpenStax College Physics, Chapter 3, Problem 61 (Problems & Exercises)
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This is College Physics Answers with Shaun Dychko. A sandal is dropped from the top of a 15 meters high mast on a ship, and the ship itself is moving at 1.75 meters per second due South. I’ve drawn our coordinate system a little bit different than usual because we want to have the perspective of somebody standing on the shore, looking at the ship. So imagine that you’re standing on the shore looking towards the West, so an arrow going into the page in which case you would see just the tail feathers of the arrow, would be West so you’re looking towards the West, the West is in front of you. Or an arrow coming out of the page would be going to the East and I draw that as a dot, that’s the tip of the arrow coming towards you, so the East is behind you, and to the right is North and to the left is South. So you’re standing on the coast of British Columbia, say looking at the Pacific across towards Japan and Alaska will be to the North on your right and California will be to the South on your left. Then you’re seeing a ship in front of you travel South. So this sandal is going to fall straight down with respect to somebody on the boat but according to an observer standing on the shore, the sandal will go down and a bit to the left from their perspective as shown here. But first let’s consider the perspective of the boat, somebody on the boat. We’re going to find out what the sandal’s final speed will be just as it’s about to hit the bottom of the boat. It starts at an initial height of 15 meters and ends at a final height of zero. It’s just dropped so the velocity of the sandal with respect to the boat, S for sandal, B for boat, naught because this is the initial velocity, is zero. Its acceleration will be that due to gravity, so negative 9.8 meters per second squared. Then we can use this kinematic formula to find its final velocity just as it hits the deck of the boat. So the final velocity squared equals initial velocity squared plus two times the acceleration times its change in height. The initial velocity is zero and the final position is zero, so this formula reduces to this where we have taken the square root of both sides to solve for vSB. So that’s the square root of two times acceleration times the negative of the initial height. So it’s square root of two times negative 9.8 meters per second squared times negative 15 meters, giving it 17.1 meters per second will be the velocity of the sandal with respect to the boat when it hits the deck. Part b says give an answer to the same question, but from the perspective of somebody standing on the shore who sees the boat also travelling to the South at 1.75 meters per second. So that’s the velocity of the boat with respect to the Earth here and our job is to find out what is the velocity of the sandal with respect to the Earth. As always for relative velocity questions, we can write down the vector addition using subscripts such that the inner subscripts are the same so that they cancel and then we’re left with our answer here having subscripts that will be the outer subscripts. So the velocity of the ship with respect to the boat plus the velocity of the boat with respect to the Earth will result in the velocity of the ship with respect to the Earth, the outer subscripts. And that’s what we want to find, the velocity of the ship, sorry the velocity of the sandal, the S stands for sandal, not ship. The velocity of the sandal with respect to the Earth. Now they’ve made things a little bit convenient here because the components of the velocity of the sandal with respect to the Earth are in the y direction completely the velocity of the sandal with respect to the boat and then the x direction, the velocity of the boat with respect to the Earth. Because each of these vectors has components in only one direction. So to find the magnitude of the velocity of the sandal with respect to the Earth, we take the square root of the sum of the squares of each of these velocities. So that’s the square root of 17.146 meters per second squared plus 1.75 meters per second squared and this gives us 17.2 meters per second. Then to find the angle of the velocity of the sandal with respect to the Earth, we’re going to measure it with respect to the horizontal just because that’s traditionally the reference point for angles is the horizontal line, I mean you could find the angle in here, that would be equally correct but I’m going to do it out here. So by taking the inverse tangent of the y component divided by the x component of this velocity of the ship with respect to the Earth, we’re finding this angle here. So that’s the inverse tangent of 17.146 meters per second y component, divided by 1.75 meters per second x component, that gives 84.2 degrees and that is below the horizontal towards the South. That gives us the final answer, the velocity of the sandal with respect to the Earth is 17.2 meters per second, 84.2 degrees below horizontal to the South.
Shouldn't the both velocities be negative?
Thank you very much for your comment. I agree that part (a) should be negative since I implied that down is the negative direction when stating that acceleration due to gravity is negative. I have updated the solution. Part (b), however, I've left unchanged. The purpose of the negative sign is to indicate direction, but the direction is given explicitly, and it doesn't go along one of the axes.
Thanks again, and all the best,
what about part C?
Ah, good point amandacav123. Thank you for noticing this! I have updated the final answer to include part (c).