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90. DEF. When a particle is projected with a given velocity under the action of its own weight, its range is the horizontal distance from its initial position at which the particle strikes the ground; its time of flight is the interval occupied between its start and the instant at which it strikes the ground.
91. PROP. To find (i) the range, (ii) the time of flight and (iii) the greatest height of a particle projected from a point on a horizontal plane.
Let u velos be the initial velocity; a the angle of projection with the horizontal plane.
Draw OQ = ut feet, making an angle HOQ=a, with the horizontal line OH; draw QP= 1gt* feet vertically downwards; then P is the position of the particle after t seconds.
Produce QP to cut OH in N.
ut cos a feet. Also
NP=NQ-PQ=OQ sin a- PQ
= (ut sin a – 1gt*) feet. When P strikes the ground NP vanishes; hence, if t seconds be the time of flight,
ut sin a - įgt = 0,
2u sin a that is, either t=0, or, t=
(a sin a -1)
Hence, when Preaches the ground again at O
zu sin a
2u sin a And then, 00', [which = (ut cos a) ft.]
Xu cos a ft.
2u sin a
uo sino a 'u sin a
u sin a this is greatest when t = ; that is, at the middle of the
u sino a time of flight; and then NP=1 feet, which is the
8 greatest height.
(iii) Example. A projectile has initially 1600 velos in a direction making an angle whose sine is to with a horizontal plane ; find the time of flight, the range, and the greatest height.
Here, sin a=it and
= (1 - sino a)=N1988=10N 99
(ut sin a - £gt2) feet; that is,
(1600 x 1o xt – 16t4) feet. This is zero when
i=10. Hence, the time of Hight is 10 seconds
(i). The horizontal distance after t seconds is ut cos a feet. Hence, the range is 1600 X 10 X 9949... that is, about 15918 feet (ii).
The greatest height is the vertical distance at the middle of the time of flight; that is,
(1600 x 5 x 1o - 16 x 25) feet. Hence, the greatest height is 400 feet
Find the range, the time of flight and the greatest height when a particle is projected from a point on a horizontal plane under the action only of its weight; the direction of projection making the following angles with the horizon and the initial velocity being as follows.
1. Angle, 45°; velocity, 100 velos.
10. The Martini-Henry rifle gives its bullet a muzzle velocity of 1,315 velos; find its range when it is elevated to an angle whose sine is tid, neglecting the resistance of the air.
11. The new Enfield-Martini rifle gives its bullet a muzzle-velocity of 1570 velos; find its range when it is elevated to an angle whose sine is ito, neglecting the resistance of the air.
12. Find the height of the bullet in Questions 7 and 8 at the middle of the flight.
13. The 110 ton gun gives its shot of 1800 lbs. (when using 950 lbs. of Waltham Powder) a muzzle velocity of 2100 velos; find its range on a horizontal plane when elevated to an angle whose sine is it.
14. The 110 ton gun gives a muzzle velocity of 2128 velos when using 1000 lbs. of powder; find the range on a horizontal plane when the elevation is an angle whose sine is }.
THE PARALLELOGRAM OF VELOCITIES.
92.' DEF. The resultant of two' simultaneous velocities is a velocity such that the distance due to it after any interval is equal to the resultant of the two distances due to the two simultaneous velocities.
93. The two simultaneous velocities are called the components of the resultant; and when they are mutually at right angles they are each called the resolved part of the resultant in their own direction.
94. In fact it will be found that all definitions and propositions applicable to distances may be applied to velocities (and also to any vector quantity). Thus we have
95. PROP. The parallelogram of Velocities. When two velocities are represented (in direction and magnitude) by two straight lines OA and OB, their resultant is represented (in direction and magnitude) by OR the diagonal of the parallelogram AOB.
Let a point P have (in addition to any other motion) the two simultaneous velocities represented by OA and OB.
Let OA represent u velos; let OB represent ú' velos; and let OR represent u velos.
Let O' represent the position of the moving point at the beginning of an interval t seconds.
Draw O'O" to represent the distance due to all the other motions of the moving point for the interval t seconds.
Draw O’Q parallel to 0 A to represent u't ft. and
draw QP parallel to OB to represent u't ft. Then P is the position of the moving point at the end of t seconds; O"Q is the distance due to the velocity OA; QP the distance due to the velocity OB; therefore OP is the distance due to their resultant velocity.
0" Q OA OA
QP u" OBOR
Hence O'P represents ut ft.; that is, O'P is the distance due to the velocity OR. In other words OR represents the resultant of the velocities OA and OB. Q. E. D.
96. When the moving point has the two velocities OA and OB only, then in Art. 95, 00” disappears, and we see that since O'P is always parallel to OR (which is fixed in direction and magnitude) and always represents ut ft., the actual motion of the moving point is the velocity OR.
97. The actual motion of a point which is moving with any number of simultaneous uniform velocities is uniform velocity.
For its resultant velocity, given by the parallelogram of velocities, is always the same; and its resultant distance from its initial position is always parallel and proportional to this resultant velocity. Its actual path is therefore a straight line along which it moves with uniform velocity.