with the horizontal line BC. Let u be the velocity of projection. Suppose the velocity of the particle always resolved into two components, one in the direction BT, and the other vertically downwards. Now since the only force acting on the particle is its weight, which acts vertically downwards, the velocity generated in each instant is always vertical, and due to a constant acceleration g. Hence during any particular instant it is only the vertical component of the particle's velocity which is affected, and this being true for each instant, it follows that throughout the motion the vertical component of the velocity is uniformly accelerated while the component in the direction BT remains the same, and is therefore always equal to u. The vertical component of the velocity which is zero at the commencement of the motion will be denoted by gt at the end of t seconds.

Now since the particle moves with a constant velocity u in the direction BT, and this velocity is independent of its. vertical velocity, it will, in t seconds, on account of this component of its velocity alone have moved through a space ut. Take BT equal to ut and draw TK vertical. Then since the other component of the particle's velocity is vertical, the space passed over during each instant by the particle on account of this second component of its velocity is always vertical, and this being true for each instant during its flight, it follows that the space passed over by the particle during any time top account of the second component of its velocity only is in the vertical direction. Hence at the end of the time 7 the particle must lie somewhere in the straight line BK. Let be its position. Now the distance TQ will be equal to the space passed over by the particle in the time t, on account of the vertical component of its velocity. But this component at the end of any time t is equal to gt, hence the space passed over from rest by the particle in t seconds moving with this component of its velocity only is gt.

1

1

Therefore TQ=gt. Draw BV vertical, and QV parallel to

1

BT. Then QV=BT= ut, and BV=.QT= 5 gt2,

G. D.

8

2u2

and is a constant quantity. But in the parabola QV4SP.PV (Besant's Conics, page 33). Therefore, the curve is a parabola whose axis is vertical and vertex upu2 wards, and the distance of B from the focus is which is therefore also its distance from the directrix.

2g

100. Assuming the path of a projectile to be a parabola, we can at once find the latus rectum and the position of the focus. The letters used to designate the equations are the same as in Art. 98.

Resolving the velocity always into two components, the first horizontal and the second vertical, these two components at the end of t seconds from the commencement of the motion are respectively u cos a and u sin a-gt. The point is moving horizontally and is therefore at the vertex A (see figure, Article 98) of the parabola when its vertical velocity is zero, u sin a that is when t= We have then

But since the curve is a parabola if I be the latus rectum, we must have

The range is equal to twice BK, since the parabola is symmetrical about AK. Hence if r be the range,

In order that the range may be a maximum we must have sin 2x=1, and therefore a=45°. Hence for a given velocity of projection an elevation of 45° gives the greatest horizontal range.

101. The directrix being at a height above A equal to one-fourth of the latus rectum, its height above BC is

1

Now the height of Q above BC at the end of the time t is usina.t-gt. Hence the distance of Q below the directrix is

Also the velocity of the particle is the resultant of its horizontal velocity u cos a, and its vertical velocity u sin a ― gt, and these are at right angles. Hence if v denote the velocity of the projectile at the end of the time t,

v=

=

{u2 cosa +(u sin a- gt)2}}

{u2 — 2u sin a . gt+g2 t2 } $ .............

(G).

Now the distance, S, of Q below the directrix is given by

and the velocity, V, of a particle falling freely through this vertical height is given by the equation

V1 = 2g8;

.. V2 = u2 - 2u sin a. gt + g3t2.

Hence the velocity of the projectilè at any time is that due to falling from the directrix to its position at the time. This is in accordance with the principle, that the increase of the kinetic energy of the body is equivalent to the work done by gravity upon it.

102. DEF. The angle which the direction of projection makes with the horizontal plane through the point of projection is called the elevation of projection.

We will now apply the methods and results of the preceding articles to some examples.

A bullet is projected with a velocity of 1000 feet per second at an elevation of 15°. Find its range on the horizontal plane through the point of projection, neglecting the resistance of the air.

The range r is given by the equation

The range is therefore 5208 yards. Of course in the case of bodies moving with such great velocities as 1000 feet per second, the resistance of the air is very great indeed, and the above result is practically worthless. The effect of the resistance of the air upon cannon shot is such, that the maximum range is obtained when the elevation of the line of fire is about 30°.

103. A particle projected at an elevation a, and with velocity u, strikes a fixed vertical plane perpendicular to the vertical plane of projection, and at a distance k from the point of projection. Find the height at which it strikes.

The horizontal velocity of the particle remains constant and equal to u cosa. Hence if t be the time elapsing after

the instant of projection and before the particle reaches the plane, we have

k

u cos a

The height to which the particle will rise in this time is given by the equation

104.

Suppose it required to find the point where a projectile will strike a horizontal plane at a distance h below the point of projection.

Let, as before, u be the velocity and a the elevation of projection. The time t in which the particle will rise to a height s above the point of projection is given by the equation