Consider the motion of a particle which is projected in a direction making an angle α with the horizon. When we neglect drag, the only force which acts upon the particle is its weight, mg (Fig. 66).

Taking the plane of motion to be the xy-plane, and applying Newton’s laws of motion gives us the equations

x-axis

(1)

y-axis

(2)

where and are the components of the acceleration along the x and y axes. Integrating equations (1) and (2) we get

Therefore the component of the velocity along the x-axis remains constant, while the component along the y-axis changes uniformly. Let v₀ be the initial velocity of the projection, then when t = 0, ẋ₀ = v₀ cos α and ẏ₀ = v₀ sin α. Making these substitutions in the last two equations we obtain

Therefore

(3)

(4)

Then the total velocity at any instant is

and makes an angle 𝜃 with the horizon defined by

Integrating equations (3) and (4) we obtain

But when t = 0, x = y = 0, therefore c₃ = c₄ = 0, and consequently

(5)

(6)

It is interesting to note that the motions in the two directions are independent. The gravitational acceleration does not affect the constant velocity along the x-axis, while the motion along the y-axis is the same as if the body were dropped vertcally with an initial velocity v₀ sin α.

bf The Path - The equation of the path may be obtained by eliminating t between equations (7) and (8). This gives

(7)

which is the equation of a parabola.

bf The Time of Flight - When the projectile strikes the ground its y-coordinate is zero. Therefore substituting zero for y in equation (8) we get for the time of flight

(8)

The Range - The range, or the total horizontal distance covered by the projectile, is found by replacing t in equation (7) by the value of T in equation (10), or by letting y = 0 in equation (9). By either method we obtain

(9)

Note that a basic trigonometric identity was used to simpilfy the above equation.

Since v₀ and g are constants the value of R depends upon α. It is evident from equation (11) that R is maximum when sin 2α = 1, or when α = . The maximum range is, therefore,

(10)

In actual practice the angle of elevation which gives the maximum range is smaller on account of the resistance of the air.

The Highest Point - At the highest point ẏ = 0. Therefore substituting this value of ẏ in equation (4) we obtain or T for the time taken to reach the highest point. Subsituting this value of the time in equation (8) we get for the maximum elevation

(11)

The Range for a Sloping Ground - Let β be the angle which the ground makes with the horizon. Then the range is the distance OP, Fig. 67, where P is the point where the projectile strikes the sloping ground. The equation of the line OP is

(12)

Eliminating y between equations (14) and (9) we obtain the x-coordinate of the point,

But x_p = R′ cos β, where R′ = OP.

Therefore

(13)

Thus for a given value of β, R′ is maximum when sin = 1, that is, when α = + .

(14)

When β = 0 equations (15) and (16) reduce to equations (12) and (13), as they should.