4. What is the velocity, per second, acquired by a heavy body in falling 10 seconds?

Ans. 320 feet.

5. What is the velocity acquired by a body in falling 25 Ans. 800 feet.

seconds?

To find the time of descent :

Divide twice the height by the acquired velocity: Or, Divide the height by 16, and extract the square root of the quotient : Or, Divide the acquired velocity by 32.

6. In how many seconds would a body fall 27,000 feet? Ans. 41.07 seconds.

7. In what time would a falling body acquire a velocity of 900 feet? Ans. 28.12 seconds.

To find the height :

Divide the square of the acquired velocity by 64: Or, Multiply the acquired velocity by half the time.

8. Required the height through which a body must fall to acquire a velocity of 1500 feet per second.

Ans. 35,156 feet.

9. A shell is projected from a mortar in a perpendicular direction with a velocity of 1,400 feet per second; required its greatest altitude, and its time of flight.

1

Ans. 30,625 feet, and 87.5 seconds.

To solve the above examples by the sliding rule :Place 16 on Cover 1 on D, and over the time, in seconds, found on D, will be found the height in feet on C; and under the height in feet found on C will be found the time, in seconds, on D. Or, (because a heavy body will fall one mile in 18.17 seconds,) Place 1 on C over 18.17 on D, and over any num.ber of seconds found on D, will be found the height in miles on C; or under any number of miles found on C, will be found the time in seconds, on D.

10. In how many seconds will a heavy body fall 44 miles ? Ans. 384, nearly.

73. PROJECTILES.

The range of a projectile is the distance from the point where it is discharged to that where it strikes the earth, measured on the earth's surface, or the horizontal distance between said points.

All projectiles moving in a vacuum describe a true parabolic curve; and dense bodies, as iron and lead balls, which do not move more than 1,000 feet per second, at the instant of discharge, describe a curve through the air which approaches very nearly to the parabolic curve.

AB represents the range, AE the elevation, (BAE being the angle of elevation,) and CD the height, sometimes called the greatest height, or greatest altitude, and likewise the height of projection. The curve ACB represents the line described by the body projected. ON represents the height of the projectile corresponding to a given horizontal distance, as AN; and EO is the variation of the projectile from AE, the tangent, or line of projection. [One of the most perfect instances of the parabolic curve is witnessed in columns of water flowing from vessels or from jets, &c., and the rules given in this paragraph, if applied to water-jets, will give very accurate results.] In consequence of the resistance of the air, the path of a body, as has been stated, deviates somewhat from a true parabola into a curve called the ballistic The greatest range of any projectile on a plane (its impetus or velocity of projection being given) is when it is discharged at an angle of 45 degrees; in which case the height to which it will rise, is just half that to which it would attain, if it were discharged with the same velocity in a vertical direction ; and its range is just twice the distance to which it would ascend, if it were projected with the same velocity in a vertical direction; and the time of its flight is as much less than it would be, were it discharged in a vertical direction, as the square root of .5, viz. 0.7071065, is less than 1. Therefore,

To find the range of a projectile, its greatest altitude, and its time of flight, when discharged with a given velocity at an angle of 45 degrees :

Divide the square of the velocity of projection by 32, and the quotient will be the RANGE. Divide the RANGE by 4, and the quotient will be the greatest ALTITUDE:-Divide the square root of the RANGE by 4, and the quotient will be the TIME of FLIGHT :-Or, (because a range of 16 feet is passed over in 1 second of time,) Place 16 on Cover 1 on D, and under the RANGE found on C, will be found the TIME of FLIGHT on D; or over the TIME OF FLIGHT found on D, will be found the RANGE on C.

EXAMPLES.

1. A shell was discharged with a velocity of 1,000 feet per second, the angle of elevation being 45 degrees; required the height to which it would ascend if projected with the same velocity in a vertical direction, its greatest altitude, its range, and its time of flight.

Answers in order, -15,625 feet; 7,812.5 feet; 31,250 feet; and 44.26 seconds.

2. A ball was discharged from a cannon with a velocity of 1,500 feet per second, the gun being elevated at an angle of 45 degrees; required the range, the greatest altitude, and the time of flight.

Ans. in order,-70,312.5 feet, or 13.314 miles; altitude, 3.328 miles; time, 66.29 seconds.

3. The greatest altitude of a shell was 1,600 feet, (the angle of elevation being 45 degrees ;) required its range and time of flight. Ans. Range 6,400 feet; time 20 seconds.

4. The range of a shot discharged at an angle of 45 degrees, was 12,000 feet; required its greatest altitude, and its velocity of projection. Ans. Range 3,000 feet; velocity 617.67 ft.

The range of a projectile, discharged with the same velocity, the angle of elevation being the same number of degrees either above or below 45, is the same; and when discharged at an elevation of 30 degrees above or 30 below 45, the range will be only half as great as the range at 45 degrees.

5. What will be the range of a shell discharged from a mortar with a velocity of 579 feet per second, the angle of elevation being either 15 or 75 degrees ? Ans. 5,238 feet.

To find the range of a projectile discharged at any given angle of elevation :

elevation by the NATURAL (The sine of any angle of

Multiply the range at 45 degrees SINE of twice the angle of elevation. more than 90 degrees is the same as the sine of its supplement. See 63, page 181.)

To find the time of flight for any elevation :

Divide the velocity of projection by 16, and multiply the quotient by the sine of the angle of elevation.

To find the greatest altitude :

Multiply half the range at 45 degrees by the SQUARE of the sine of the angle of elevation.

The sine of one second is .0000048481, very nearly; and the sine of any number of seconds from 1 to 60 may be found very nearly by multiplying the sine of one second by the number of seconds. The sine of one minute is .000290888, very nearly; and the sine for any number of minutes from 1 to 60 may be found nearly by multiplying the sine of 1 minute by the number of minutes.

The following table shows the sine of each degree from 1 to 90 inclusive:ー

NATURAL SINES.

Deg. Sines. Deg. Sines. Deg. Sines. Deg. Sines. Deg. Sines.

1.017452 19.325568 37.601815 55.819152 73.956305 2.034899 20.342020 38.615661 56.829038 74.961262 3.052336 21.358368 39.629320 57.838671 75 .965926 4.069756 22.374607 40.642788 58.848048 76.970296 5.087156 23.390731 41.656059 59.857167 77.974370 6.104528 24.406737 42.669131 60.866025 78.978148

7.121869 25.422618 43.681998 61.874620 79 .981627 8.139173 26.438371 44.694658 62.882948 80.984808 9.156434 27.453990 45.707107 63.891007 81.987688 10.173648 28.469472 46.719340 64.898794 82 .990268 11 .190809 29.484810 47.731354 65.906308 83.992546 12 .207912 30.500000 48.743145 66.913545 84.994522

13.224951 31.515038 49.754710 67.920505 85.996195 14.241922 32.529919 50.766044 68.927184 86.997564 15.258819 33.544639 51.777146 69.933580 87

.998637

16.275637 34.559193 521.788011 70.939693 88

.999391

17.292372 35.573576 53.798636 71.945519 89

.999848

18.309017 36.587785 54.809017 72.951057 90 1.000000

The natural sines for degrees and parts of a degree, as that for 10 degrees and 20 minutes, may be found very nearly as follows:-Subtract the natural sine of the given number of degrees from the next greater; and say:-As 60 is to the remainder, so is the number of minutes to the part to be added to the sine of the given number of degrees. Thus, .190809.173648 = .017161. Then, as 60: 017161 :: 20 to the part to be added to the sine of 10 degrees, viz. 0.005720; consequently, the sine of 10 degrees and 20 minutes = .179368, nearly.

6. A cannon ball is fired with a velocity of 480 feet per second, at an elevation of 75 degrees, and another is fired with the same velocity at an angle of 15 degrees; required the ranges. Ans. 3,600 ft., nearly.

7. If a shell be fired with a velocity of projection sufficient to give it a range of 3,600 feet at an elevation of 45 degrees; required its range and time of flight, when discharged with the same velocity at an angle of 32 degrees.