The cannon on a battleship can fire a shell a maximum distance of 32.0 km. (a) Calculate the initial velocity of the shell. (b) What maximum height does it reach? (At its highest, the shell is above 60% of the atmosphere—but air resistance is not really negligible as assumed to make this problem easier.) (c) The ocean is not flat, because the Earth is curved. Assume that the radius of the Earth is $6.37 \times 10^3 \textrm{ km}$. How many meters lower will its surface be 32.0 km from the ship along a horizontal line parallel to the surface at the ship? Does your answer imply that error introduced by the assumption of a flat Earth in projectile motion is significant here?

a) $560 \textrm{ m/s}$

b) $8.00 \textrm{ km}$

c) The Earth's surface will be $80 \textrm{ m}$ below the horizontal line parallel to the surface at the ship. The assumption of a flat Earth is acceptable for a range of $32 \textrm{ km}$ since this discrepancy represents only $1%$ of the maximum height of the projectile.

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*theta*equal to 90, you get the maximum angle for the maximum range. That'll happen when

*theta*is 45. So our max happens when

*theta*is 45. So that means we can replace this sine two

*theta*which is sine of 90 in other words or one, with -- you know it disappears because it's just the number one. So we have

*R max*is the initial launch speed squared divided by the acceleration of gravity and then we'll solve this for

*v naught*. We'll multiply both sides by

*g*and then we get

*v naught*squared is

*g*times

*R max*and then take the square root of both sides, and you get

*v naught*is the positive square root of

*g*times

*R max*. That is the square root of 9.8 meters per second squared, times 32000 meters which is 560 meters per second. Notice that I had to convert this range into meters because kilometers would not go with the meters that are in this number and so we converted this range of 32 kilometers by multiplying by 1000 meters per kilometer, giving us 32000 meters. Okay. So then, the next question is what maximum height will the projectile reach? This formula is going to help us with that because we know some things about the projectile when it's at its maximum height. The velocity in the y direction will be zero at the very top of its trajectory, right here. We know what the initial launch height is, it starts at zero, that's what we defined here,

*y naught*equals zero. So then this equation reduces to two

*ay*equals negative

*V naught y*squared where we've subtracted

*v naught y*squared from both sides and then switch the sides around. So we have two times acceleration due to gravity times the maximum height equals negative of

*v naught y*squared, the

*y*component of the velocity squared. Then divide both sides by two

*a*and we get the maximum height is negative of

*v naught y*squared over two

*a*and the

*y*component of the launch velocity is

*v naught*times sine of the launch angle. So that gives us negative of 560 meters per second which we calculated in part A as the launch velocity, times sine of the launch angle that gives the maximum range of 45 degrees and square all that. The negative by the way is outside of the square so this numerator remains negative. And divide by two times negative 9.8 meters per second squared, giving us 8 kilometers as the maximum height. Now in part C, we've made an assumption with our range formula that the height of the, where the projectile lands is the same as the height where it was launched from. It assumes the earth is flat in other words. So now, let's check to see whether that assumption is reasonable. We're going to be having to find this little bit of distance here between the tangent which is at the point where the battleship is and so this is the assumed, right here on the tangent line is the assumed ground or water where the shell hits the water. Down here a little bit is where the curved surface of the earth actually is. This discrepancy is

*delta y*is the amount we're going to try and find in part C and we're going to see if it is large or if it is insignificant. So the distance between the point on the earth here where the shell lands and the point where it was launched is this arc length along the curve of the earth and that is 32.0 kilometers. So what we have here is a triangle. These are the sides of the triangle and we're going to find -- we know what this leg is here, this is the radius of the earth between the battleship and the center of the earth. We're going to calculate what the hypotenuse length is and subtract away the radius of the earth and that will leave us with this portion

*delta y*. We don't care what this leg here of the triangle is. So in order to figure out what the length of this hypotenuse is we need to know what this angle is, which we can find because we know that the arc length is 32 kilometers and we'll see what fraction of a full circle that arc length is. So we'll divide the arc length by a full circle circumference and then multiply by 360 degrees per circle and that will give us the number of degrees of this angle here. So we have 32 kilometers divided by two pi times the radius of the earth, 6.37 times ten to the three kilometers, and that gives us what fraction of a circle this arc covers, times it by 360 degrees per circle and you get 0.28782 degrees. Here I've re-drawn this triangle that I've highlighted in purple, re-drawn it here, and it's a right triangle and it has one side radius of the earth as

*theta*which we just calculated and it has the hypotenuse which is the radius of the earth plus this discrepancy. Yeah. So cos

*theta*is the adjacent radius of the earth divided by the hypotenuse, radius of the earth plus the

*delta y*. We'll multiply both sides by radius of the earth plus

*delta y*and then we get this line here. Then we'll divide both sides by cos

*theta*and we get radius of the earth plus the

*delta Y*equals radius of the earth divided by cos i and then subtract radius of the earth from both sides and we get

*delta y*. So that's going to be radius of the earth divided by cosine of 0.28782, minus radius of the earth giving us an 80 meter discrepancy which is not significant because it represents only one percent of the maximum height of the projectile.