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The cannon on a battleship can fire a shell a maximum distance of 36.0 km.
(a) Calculate the initial velocity of the shell.
m/s

(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.)
m

(c) The ocean is not flat, since the earth is curved. How many meters lower will its surface be 36.0 km from the ship along a horizontal line parallel to the surface at the ship?
mDoes your answer imply that error introduced by the assumption of a flat earth in projectile motion is significant here? (Select all that apply.)
The error is insignificant compared to the distance of travel.
The error is significant compared to the distance of travel.
The error could be significant compared to the size of a target.
The error is insignificant compared to the size of a target.

Sagot :

(a)The initial velocity of the shell will be 594.27 m/sec

(b)The maximum height it reaches will be 9000 m.

c)101.249 m meters lower will its surface be 36.0 km from the ship along a horizontal line parallel to the surface of the ship.

d)The error could be significant compared to the size of a target. Option C is correct.

What is velocity?

The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

The given data in the problem is

m is the mass of the block = Kg.

u is the initial velocity of fall = m/sec

h is the distance of fall =  m

g is the acceleration of free fall = m/sec²

v is the hitting velocity of =?

a)

The range of the projectile is;

[tex]\rm R = \frac{u^2 sin 2 \theta }{g} \\\\ 36 \times 10^ 3 = \frac{U^2 sin 45^0}{9.81} \\\\ U= 594.27 \ m/sec[/tex]

b)

The maximum height of the projectile is;

[tex]\rm H = \frac{u^2 sin 2 \theta }{2g} \\\\ H = \frac{(594.27)^2\times (sin 45)^2}{2 \times 9.81 } \\\\ H = 9000 \ m[/tex]

c)

The distance between its surface and the ship, measured in a horizontal arc parallel to the surface, will be 36.0 kilometers. The distance from the lower surface is found as;

[tex]\rm( R_e + h)^2 = R_e^2+(36)^2 \\\\ (R_e)^2 = h^2+2R_e h= R_e^2 + 12196 \\\\ h^2 + 12800 h - 1296 = 0 \\\\ h = 101.249 \ m[/tex]

d)

An error is a mistaken or erroneous action. In some contexts, an error is interchangeable with a mistake.

The difference between the calculated value and the original value is known as the error. The inaccuracy may be large in comparison to the target's size. Option C is correct.

Hence the initial velocity of the shell, maximum height, and the distance from the lower surface will be 594.27 m/sec,9000 m, and 101.249 m and option c for question d are correct respectively.

To learn more about the velocity, refer to the link: https://brainly.com/question/862972.

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