say you dropped a cannonball from the 17.0–meter mast of a ship sailing at 2.0 meters/second. how far from the base of the mast will the ball land? a. 0 meters b. 2 meters c. 2.5 meters d. 3 meters

The correct choice is A. 0 meters.

If you simply dropped the cannonball and didn't throw it horizontally,

then it'll fall straight down the mast and land on the deck right next to

the mast.

While you were up there holding it, before you dropped it, the cannonball

was moving horizontally, at 2.0 meters/second, along with the rest of the

ship and everything else aboard. It continued doing that after it dropped,

and from the point of view (in the reference frame) of the mast and everyone

on the ship, it fell straight down, parallel to the mast.

Now that we have that question answered, we can proceed to the more-

important ones. I answered the easy one, but YOU'll have to answer these:

==> WHY did you climb the mast carrying a cannonball ?

Have you been drinking sea water or bad rum ?

==> WHY did you drop it, and never even yell "LOOK OUT BELOW !" ?

==> How many formerly-able-bodied souls were injured by the

plummeting cannonball ?

==> What did everybody ELSE yell after the impact ?

==> What did they do to you after they brought you down ?

A). 0 Meter

Explanation:

As we know that cannon ball is dropped downwards from ship of height 17 m

so here the time taken by the ball to drop it downwards is given by

now in the same time the ball will travel horizontally with the speed same as the speed of the ship

so here we will have

so it will move by 3.72 m from its initial position

but in the same time mast of the ship will also cover same distance as ship is also moving with same speed

So the distance between the base of ship and the ball will be

d = 0 m

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j

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