9 times 16.1 in three seconds; 16 times 16.1 in four seconds, and so on; the whole space fallen through in any number of seconds, being equal to the square of the time in seconds multiplied by 16.1 feet. Now a heavy body, as a ball of iron or of lead, in falling through the atmosphere, is governed by the same law, only its velocity is slightly impeded by the resistance of the atmosphere, for which it is proper to allow onetenth of a foot for every 16 feet. Hence, To find the distance fallen through by any heavy body in a given time: Multiply the square of the time in seconds by 16, and the result will be the distance in feet. EXAMPLES. 1. A man fell from a balloon, and was 25 seconds in reaching the ground; required the height of the balloon. Ans. 10,000 feet. 2. A shell fired from a cannon, after having reached its greatest height was 20 seconds in falling to the earth; required the height which it reached above the earth. Ans. 6,400 feet. Now, since bodies ascending lose equal velocities in equal times, a body projected upwards will lose all its motion in the same time required to fall back to the earth. Consequently, the shell which was 20 seconds in falling must have been 40 seconds in the air; its time of flight being 40 seconds. 3. A meteor is seen to explode in the air, and 50 seconds after the fragments reached the earth; required its height at the time of the explosion. Ans. 7.57575 miles. A body in falling 1 second acquired a velocity of 32.2 feet, or, if we allow for the resistance of the atmosphere, a velocity of 32 feet; and the velocity gained after the first second is directly as the time. Therefore, To find the velocity acquired by a falling body:- 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 seconds? Ans. 800 feet. 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. Ans. 30,625 feet, and 87.5 seconds. To solve the above examples by the sliding rule :— Place 16 on C over 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 number 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 4 miles? Ans. 381, nearly. T73. 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 curve. 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 C over 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 : Multiply the range at 45 degrees SINE of twice the angle of elevation. more than 90 degrees is the same as See63, page 181.) elevation by the NATURAL (The sine of any angle of the sine of its supplement. 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 : |