Object projected at an angle

Consider now the case of an object that is projected at an angle to the horizontal other than 90^o.
It is helpful to treat the horizontal and vertical components of velocity separately. A diagram of the motion is shown in Figure 1.
Consider an object projected with velocity u at an angle A to the horizontal.

Vertical component of velocity = a sin A
Horizontal component of velocity = u cos A

If we ignore the effects of air resistance, the horizontal velocity is constant and the vertical velocity changes with a uniform acceleration. The path that the body follows is a parabola as can be seen from the proof below.

Vertical motion: h = ut sin A – ½ gt²

Horizontal motion: s = ut cos A

(a) Range

The object will hit the ground again when h = 0, i.e. when ut sin A = ½ gt².
Therefore it will hit the ground after a time t, where t = 2u sin A/g.

Therefore the range R is given by:
R = horizontal velocity x time = [ucos A x 2u sin A]/g = [u² 2 sin A cos A]/g = u²sin 2A/g


The maximum range for a given velocity of projection is when sin 2A = 1, that is, when 2A = 90^o or when A = 45^o

(b) Height

The projectile will reach its maximum height when the vertical component of its velocity is zero, that is, when:
u sin A - gt = 0, or t = u sin [A/g]

This gives the maximum height (H) reached as:
H = [u²sin² A]/2g

Trajectory – parabolic path

The shape of the trajectory can be found by combining the equations for vertical and horizontal velocity.
Taking the vertical displacement as y and the horizontal displacement as x, we have.
x = ut cos A
y = ut sin A – ½ gt ²

This gives: y = x tanA – gx²/²u² cos²A

For a given angle of projection (A) and projection velocity (u), this becomes:
y = Bx - Cx²
where B and C are constants (B = tanA and C = g/2u² cos²A) ; this is the equation of a parabola.

The above equations only refer to projectiles where the effects of air resistance have been ignored. Quite different paths may be found in practice for such objects as golf balls, javelins and discoi.