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 x-direction is the same amount of time that it takes to travel in the y-direction. Since in the Y-direction 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 x-direction, use D=VT and solve for the distance in the x-direction using the value for T that you calculated. D=(4.63m/s)*T.

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