The cannon on a battleship can fire a shell a maximum distance of 25.0 km. (a) C

Question

The cannon on a battleship can fire a shell a maximum distance of 25.0 km. (a) Calculate the initial velocity of the shell. (b) What maximum height does it reach? (At its highest, the shell is above a substantial part of the atmosphere--but air resistance is not really negligible as assumed to make this problem easier.) (c) The ocean is not flat, since the earth is curved. How many meters lower will its surface be 25.0 km from the ship along a horizontal line parallel to the surface at the ship?

Explanation / Answer

(a) Calculate the initial velocity of the shell. (in m/s)        

The cannon's range depends upon the angle of the barrel (known as elevation).              

I'm going to assume an elevation of ? = 45° since tan 45° = 1. This means that

for each meter of height the shell rises, it also moves downrange a meter of

horizontal distance.       

R = horizontal range = 25.0           km = 25,000        m

? = angle of the barrel = 45          °

g = gravitational acceleration = 9.81         m/s²

v = initial velocity of the cannon shell = to be determined            

R = (v²sin 2?)/g

Rg = v²sin 2?      

Rg/sin 2? = v²    

v² = Rg/sin 2?    

v = sqrt(Rg/sin 2?)          

v = sqrt[(25,000 m)(9.81 m/s²)/sin 2(45°)]            

v = 495.23 m/s  

(b) What maximum height does it reach? (in m)               

H = maximum height reached by the cannon shell =        to be determined           

H = (v²sin² ?)/2g              

H = [(495.23 m/s)²sin² 45°]/2(9.81 m/s²)

H = 6250 m         

(c) The ocean is not flat, since the earth is curved. How many meters lower will

its surface be 25.0 km from the ship along a horizontal line parallel to the surface

at the ship?       

r = radius of the earth = 6,371     km = 6.37E+06   m

Average curvature of the earth = 7.98    in /         1              mi

Average curvature of the earth = 20.2692             cm / 1.6093         km

Average curvature of the earth = 12.6    cm/km

So, over a distance of 28.0 km the earth surface drops

25.0 km x 12.6 cm/km = 315         cm = 3.15 m

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.