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127. PROP. The speed of a projectile at any point of its path is equal to that due to a fall from the directrix.

Let S be the focus and P the point of projection, then, SP is the vertical distance of P from the directrix; and

u by the last Article, SP = ft.

28 Let h be the number of ft. in SP; then, the velocity v due to falling with acceleration g, a distance h ft. is given by

v2 = 2gh but from above

uo = 2gh. Thus the speed of projection is equal to that due to falling from the directrix.

And if this is true at the point of projection it must be true at every point; for the particle might be projected from any point in its path.

NOTE. This proposition may be stated as follows; When any particle has fallen vertically under the action of gravity from a point on the directrix to a point on the parabola it has a vertical velocity downwards, whose magnitude is equal to the velocity which the particle describing the parabola has when at that point. It must be noticed that these two velocities are not in the same direction.

It follows that the horizontal speed is that due to falling from the directrix to the vertex, and hence that the number of feet in the

2u2 cosa latus-rectum of the parabola is

128. Another proof of Art. 127 is as follows.

Let the tangent at P cut the axis of the parabola in T. The vertical velocity at A is zero; so that the vertical velocity at. P (u sin a) is that due to falling a vertical distance AN; hence usina=28 AN. [Now in the parabola with the usual notation

2AN=NT=PT sin a= 2PY sin a=2SP sinoa), whence

usin’a=2g SP sinoa,

uo=2g SP.



Q. E. D.

Example i. To find the range and time of Aight on an inclined plane, angle a, which passes through the point of projection.

Resolve the velocity of projection along and perpendicular to the inclined plane; the resolved parts are, in velos, initially,

u cos (0 - a); u sin (0 - a);
u cos (0-a) – g sin a xt; u sin (0 –a) - gcos a xt.

after t secs,

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The distances resolved in the same directions after t secs., are in ft.,

u cos (0 - a) t- g sin ata; u sin (0 - a) t- 1 g cos ata. The particle strikes the plane again when the distance perpendicular to the plane viz. u sin (0 - a) t- 1 g cosata is zero; that is, when

zu sin (0 – a)

& cos a Let 00" in the figure be the range ; draw O'M perpendicular to the horizontal line OO'; then OM=horizontal velocity x time of flight

zu sin (0 - a) =u cos Ox

gcos a

OM also,



2u2 sin (0 a) cos e that is, the range is

feet. 8

cos'a Example iï. Prove that in Example i. the particle at the middle of the time of flight is vertically above the middle point of the range and is moving parallel to the inclined plane.

The first statement is most easily proved by observing that if the projectile carries a fine thread which hangs always vertical, this thread travels with uniform velocity in a horizontal direction; and that therefore the point in which this vertical thread cuts the inclined plane travels with uniform velocity along the plane.

To prove the second statement, we observe that the velocity of the projectile perpendicular to the inclined plane is zero, when

u sin (a – B) g cos B x t is zero,

u sin (a - b) that is, when


g cos B which is at the middle of the time of Alight. Therefore at the middle instant of the time of flight the velocity of the projectile is parallel to the plane.




1. A train is moving at the rate of 45 miles an hour, when a ball is dropped from the roof inside one of the carriages; prove that the ball describes a parabola in space, and find the position of the axis and directrix,

2. A boy throws a stone from the top of a cliff 128 ft. high with 64 velos at an angle 30° to the horizon; find how far from the foot of the cliff it will strike the sea and with what velocity.

3. Prove that, neglecting the resistance of the air, the range of a projectile on a horizontal plane is greatest when the angle of projection is 45°.

4. Find with what velocity a man must be able to throw that he may just be able to throw a cricket ball 100 yards.

5. A man throws a ball with a velocity 50 velos; in what direction must he throw it that he may just strike the top of a pole 40 ft. high at a distance of 100 ft. from him?

6. A particle is projected with velocity u velos at an elevation a; find its least velocity, and find when and where it attains that velocity.

7. A particle is describing a parabola under the action of gravity, and at the instant when it has its least velocity that velocity is doubled; shew that the time of reaching the horizontal plane from which it was projected is unaltered.

8. From the top of a tower I project particles in a horizontal direction; prove that the distances, from the foot of the tower at which the particles strike the horizontal plane on which it stands, are proportional to the velocities of projection.

9. A projectile has initially u velos and its angle of projection is a; prove that its distance from the point of projection after t seconds is

(u-f- - u sin a gt3 + + 3244) feet. 10. Find the angle of elevation that the horizontal range may be equal to the distance of the point of projection from the directrix.

11. Shew that the greatest range on an inclined plane of 30°, two-thirds of the greatest range on a horizontal plane, with the same initial velocity.

12. A straight smooth tube AB, 72 ft. long, is placed at an angle 30° to the horizon so that its lower end B is at a height 16 ft. from a horizontal plane; a particle is allowed to slide from A through the tube and then to describe a parabola freely; find how long after leaving the tube the particle strikes the ground.

13. In Question 12, find how far from the point vertically under B is the point at which the particle strikes the ground.

14. A straight smooth tube AB, a ft. long, inclined at an angle a to the horizon is fixed so that the lower end B is h ft. from a horizontal plane. A particle slides from A through the tube and then describes a parabola freely, striking the plane at C; find the time of falling from B to C and the distance BC.

15. A man in a railway carriage, moving with uniform velocity 30 miles per hour, throws a ball so that it goes up four feet and returns to the point of projection; shew that the ball describes a parabola relatively to the earth and find how far the train moves while it does so.

16. Two men 5 ft. apart in a railway carriage which is moving uniformly with a velocity of 45 miles an hour toss a ball from one to the other, projecting it so that its time of flight is į second. Shew that the ball describes a parabola in space; and taking its path in the carriage to be perpendicular to the rails, find its range in space.

17. The top of a railway carriage is 12 ft. from the ground, and it is moving at the rate of 7 miles per hour; a man jumps from the top, giving himself a horizontal velocity perpendicular to the rails of 84 velos; find the velocity with which he will reach the ground.

18. A particle is projected with velocity u at an angle a; with what velocity must I move on a horizontal plane that I may be always

Ι vertically under the projectile ?

19. A particle is projected from a point on an inclined plane of angle ß with velocity u at an angle a to the horizon so that its path is in the same plane as the line of greatest slope in the plane; with what velocity must I move on the plane so as to be always vertically under the particle ?

20. Morin's machine consists of a vertical circular cylinder which is made to rotate with uniform velocity, while one of the weights of an Atwood's machine having a pencil fixed to it descends with constant acceleration in a vertical line, and as it does so marks a line on the surface of the cylinder; shew that if the cylinder is covered with a sheet of paper, which after the pencil has made its mark upon it, is taken off and flattened, then the trace of the pencil on the paper will be a parabola.


21. Find the length of the latus rectum of the parabola described on a Morin's machine, in which the circumference of the cylinder is 3 st. and the weights in the Atwood's machine are 151 oz. and 161 oz. respectively (g=32], the cylinder making a complete revolution in 6 seconds.

22. A particle is projected with velocity u velos at an angle a; shew that it reaches the further extremity of the latus rectum in

(u sin a + u cos a) seconds.

8 23. A particle slides down a smooth inclined plane of angle a, having initially a horizontal velocity on the plane in a direction perpendicular to the line of greatest slope) of u velos; shew that the particle traces out a parabola on the plane and find its latus rectum.

24. A particle is projected with velocity u velcs at an angle a; shew that the focus is (sin’a - cosa) above the horizontal plane

28 through the point of projection.

25. The focus of the path of a projectile is above or below the horizontal plane through the point of projection according as the vertical component of the velocity of projection is greater or less than the horizontal component.

26. A particle slides down a smooth straight tube and then falls freely under the action of gravity; prove that the directrix of its parabolic path passes through the upper end of the tube.


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