Question

You are a Pirate Captain on a journey. Some of your newly recruited gunners still haven't gotten the swing of things and set all the cannons at different angles. Halfway through the journey, you spot a Royal Navy ship and have to go into battle with the crew that you have.

Along the ship's port (left) side you notice an equal number of cannons elevated at 30 degrees, 40 degrees, and 60 degrees respectively. Your cannons fire frictionless projectiles with a velocity of 400 m/s. (They are unaffected by air resistance)

a) How close to the Royal Navy ship should you sail in order to maximize the number of your port side cannons that will hit its deck? (Assume that the deck of the Royal Navy Ship is the same height above the water as yours)

b) Pretend for a moment that you are one of the cannonballs being shot out of one of the 30 degree cannons. At what angle would it appear to you that the cannonballs being shot out of the 45 and 60 degree were being shot at above the horizontal? Assume you maintain a strong sense of what constitutes "horizontal". (HINT: It may help to convert their trajectories onto vector form.)

Answer 1

a) Horizontal component of fire from cannon Vx = and Vertical component Vy =

Here V is 400 m/s, Now time taken for fire to go up to top and return at same height can be computed using Newtons First law of motion

Vf = U + at

Here Vf = - and U =   and a = -g

t = 2 /g

Distance Travelled in horizontal direction S = t*

for angle = 30 degree

for angle = 45 degree

for angle = 60 degree

Hence, Ship should be at distance less than 14.139 Km from the other ship to maximize the number of possible hits.

b)

For 45 degree cannon

Horizontal component of velocity Vx =

Horizontal component of velocity Vy =

Hence angle =

   = 127.50 degree

Similarly for 60 degree cannon

angle = -37.50 degree

answered by: anonymous