Answer:
Answered
Explanation:
u = 1.75 m/s
h = 15 m
Now, for the magnitude of velocity when hitting the deck
Using third equation of motion
v^2 - 1.75^2 = 2×9.8×15
v = 17.15 m/s
the velocity of sandal when it hits the deck is 17.15 m/s this the velocity of sandal relative to the ship
b) velocity relative to a stationary observer on shore
=[tex]\sqrt{1.75^2+17.15^2}[/tex]
=17.23 m/s
To a person aboard the ship, they see the sandal land directly below the release point. A stationary observer on the outside sees the same thing, except the fact that the landing point has translated to the south.
The velocity of the sandal
(a) relative to the ship is 17.15 m/s vertically downwards
(b) relative to the observer on the shore is 17.23 m/s
(c) The motion of the sandal differs for observers on the ship and observers on the shore as discussed below.
given information:
initial velocity of the ship u = 1.75 m/s
height of the mast h = 15 m
let v' be the final velocity of the sandal and u' be the initial velocity of the sandal.
Now, u' = 0 when it is dropped from the mast.
(a) From the third equation of motion
[tex]v'^2-u'^2=2gh[/tex]
where v is the final velocity
[tex]v'^2 = 2\times9.8\times15\\\\v' = 17.15 m/s[/tex]
the velocity of the sandal when it hits the deck is 17.15 m/s
(b) velocity relative to a stationary observer on shore
the horizontal component of velocity of the sandal = u = 1.75 m/s
and the vertical component of the velocity of the sandal = v' = 17.15 m/s
the resultant velocity:
[tex]v=\sqrt{v'^2+u^2}\\ \\v=\sqrt{(17.15)^2+(1.75)^2}\\\\ v=17.23m/s[/tex]
so, the relative velocity to an observer on the shore is 17.23 m/s
(c) For an observer on the ship, the sandal hits the deck directly below the point where it was dropped, but for an observer on the shore, the motion of the sandal is projectile motion since the sandal is also moving towards the south with the ship.
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