(18) If R, R', be the ranges of two projectiles, which, being thrown from the same place, attain the same vertical height, and pass through a common point, prove that

where H is the greatest height attained, k the height of the common point, and h the horizontal distance of the point of projection from the vertical line through the common point.

(19) A body is projected from a certain point vertically downwards: at the same moment another body is projected from another point with the same velocity as the former body find the direction of projection of the latter body in order that it may strike the former.

If a be the inclination of the distance between the two points to the horizon, the inclination of the direction of projection of the latter body to the horizon must be equal to – 2α.

(20) If V, V', V", be the velocities at three points P, Q, R, of the path of a projectile, where the inclinations to the horizon are a, a-ß, a— 2ß, and if t, t', be the times of describing PQ, QR, respectively, prove that

(21) A body is projected from a given point and strikes another given point: supposing this to be possible for only one angle of projection, determine the velocity of projection.

If b be the altitude of the latter point above the former, and c the distance between them, the required velocity is equal to

{(b+c) g}1.

(22) If a body be projected with a velocity V, in a direction making an angle a with the horizon, shew that, at a height h,

the direction of its motion will make with the horizon an angle equal to

2gh
sec2 a) 3.
V

a).

(23) If r1, r2, ", be the distances of a projectile from the point of projection, when its angular elevations above this point are respectively a,, a, a,, prove that

"1 cos2α, sin (α, -α)+r, cos2 α, sin (a,—α) +r, cos2α, sin (α, — α) = 0.

(24) A cannon is pointed in a direction making an angle of 30° with the horizontal plane on which it stands, and fired against a fort: it is then drawn of a mile nearer the fort, and pointed at the same elevation to the horizon as before, when it is observed that the ball strikes the fort in the same point as in the former case. If the greatest distance which the cannon can throw the ball is one mile, prove that the height of the point which the ball strikes is 165 feet above the horizontal plane on which the cannon stands.

(25) A body is projected, at an angle a to the horizon, so as just to clear two walls of equal height a, at a distance 2a from one another prove that the range is equal to

(26) A body is thrown over a triangle, passing from one extremity of the horizontal base, just over the vertex, to the other end of the base: prove that

where is the angle of projection, and a, ß, are the angles at the base of the triangle.

(27) It is required to throw a shell from a point at a distance a from the foot of a wall, so that it may just clear the top of the wall, the height of which is h, and strike the ground, which is horizontal, at a distance b beyond the wall. Determine

the velocity and angle of projection, neglecting the resistance of the air.

If a denote the angle and V the velocity of projection,

(28) If h1, h2, h,, be the heights of the sights of a rifle, when adapted for shooting at the distances 100, 200, and 400 yards, respectively; prove that

4h ̧ (h ̧2 — h ̧2) +2h ̧ (h ̧2 − h ̧2) + h ̧ (h ̧2 — h‚ ̧3)

= 0.

(29) A body is projected from the deck of a ship, with a velocity V, relatively to the ship, and at an angle of elevation B, so as to hit a mark in the wake of the ship and in the horizontal plane of the deck: the ship is supposed to be sailing in a straight course with a velocity V'. Shew that, if a be the angle of projection in order that the body, propelled by the same force, might have hit the same mark had the vessel been at rest, then

(30) Determine the angle of elevation at which a body must be projected in order that the focus of its path may lie in the horizontal plane passing through the point of projection.

The angle of projection must be 45o.

(31) If any number of bodies be projected in different directions from the same point with equal velocities, find the locus of the foci of the parabolas.

If V denote the velocity of projection, the locus will be a sphere, of which the centre is the point of projection and V2

radius

2g

(32) A number of heavy particles are projected in any directions in a vertical plane from the same point, (1), with the same vertical velocity, (2), with the same horizontal velocity:

prove that in each case the locus of the foci of their paths is a parabola with its focus at the point of projection and axis ✓ vertical, but that, in the first case, the vertex is upwards, and, in the second case, downwards.

(33) A particle is projected horizontally with a velocity of 32√3 feet per second: after a certain time, under the action of gravity, it has a velocity of 64 feet: find this time: determine also the spaces described, horizontally and vertically, and the latus rectum of the curve, supposing the force of gravity to be measured by 32 feet.

The required time is one second, the horizontal and vertical spaces described, 32√3 and 16 feet respectively, and the latus rectum of the parabola 192 feet.

(34) Heavy particles are projected horizontally with dif ferent velocities from the same point; shew that the extremities of the latera recta of the parabolas, which they severally describe, lie on a cone, of which the axis is vertical, and the vertical angle

2 tan1 2.

(35)

Swift of foot was Hiawatha;

He could shoot an arrow from him,
And run forward with such fleetness,
That the arrow fell behind him!

Strong of arm was Hiawatha;

He could shoot ten arrows upward,

Shoot them with such strength and swiftness,
That the tenth had left the bow-string

Ere the first to Earth had fallen.

Supposing Hiawatha to have been able to shoot an arrow every second, and, when not shooting vertically, to have aimed so that the flight of the arrow might have the longest range, prove that it would have been safe to bet long odds on him if entered for the Derby.

W. S.

19

SECT. 2. Motion resolved parallel to any straight lines.

(1) If a body be projected from a given point so as to strike an inclined plane through that point at right angles, to prove that

where is the angle which the direction of projection makes with the plane, and a the inclination of the plane to the horizon.

The component of the velocity of projection, at right angles to the plane, is Vsin 0, and the component of gravity, at right angles to the plane, is g cos a: hence, t being the time of flight,

Again, the velocity parallel to the plane being, by the hypothesis, zero at the time of impact, the projectile velocity Vcos a, parallel to the plane, must have been destroyed by the component of gravity g sin a in the time t: hence

(2) A body is projected from a given point with a given velocity to find the direction of projection in order that its path may touch a given plane.

Let A, fig. (107), be the point of projection, BC the given plane: draw AB horizontally to meet BC in B, and draw AD at right angles to BC.

Let AB=a,

ABC= ß; let V denote the velocity of projection and the inclination of this velocity to AD.

Then, since the velocity normal to BC must, by the nature of the question, be zero when the body reaches BC, we have