Question

The cannon on a battleship can fire a shell a maximum distance of 36.0 km.

Answer 1

(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;

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

b)

The maximum height of the projectile is;

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

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;

$\rm( R_e + h)^2 = R_e^2+(36)^2 \\\\ (R_e)^2 = h^2+2R_e h= R_e^2 + 12196 \\\\ h^2 + 12800 h - 1296 = 0 \\\\ h = 101.249 \ m$

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: brainly.com/question/862972.

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