Posted by Student on Monday, January 21, 2013 at 10:07pm.
A sailor climbs to the top of the mast, 12.55 m above the deck, to look for land while his ship moves steadily forward through calm waters at 4.63 m/s. Unfortunately, he drops his spyglass to the deck below.
Where does it land with respect to the base of the mast below him?

Physics  Devron, Monday, January 21, 2013 at 10:42pm
The time (T) that it takes the spyglass to go in the xdirection is the same amount of time that it takes to travel in the ydirection. Since in the Ydirection the initial velocity is 0 and acceleration is equal to gravity (9.8m/s^2), use the equation D=ViT+1/2(aT^2). Once you plug in the values, you should get a simplified equation of D=1/2(9.8m/s^2)(T)^2. rearranging it to solve for T should result in the equation looking as followed: T=[2D/(9.8m/s^2)]^1/2. Since velocity is constant in the xdirection, use D=VT and solve for the distance in the xdirection using the value for T that you calculated. D=(4.63m/s)*T.
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