When we throw an object high up in the air, the object is called a projectile, and the path it follows is called its trajectory. Once the object is released, it no longer experiences the initial force but comes under the influence of gravity and the air drag (or friction). However, in this article, we will ignore the effects of air resistance to keep things simple. A few simple examples of projectile motion are throwing a stone (at some angle from the horizon or straight overhead), dropping a stone into a ditch, launching a satellite into Earth orbit, or sending interplanetary probes.

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Equations of motion

Before we proceed further, we assume that you are familiar with the three equations of motion, namely,

v = u + f\times t  \hspace {30}-----[A]

s= ut + \frac {1}{2}ft^2        \hspace {30} -----[B]

v^2 = u^2 + 2fs    \hspace {30}        ----- [C],

where u is the initial velocity, v is the final velocity, f is the acceleration, and t is the flight time.

Physics of projectile motion

Let us draw a diagram to understand the physics of the projectile motion. Assume the object’s initial velocity is v, and it is projected at an angle from the horizon. Also, assume that it attains a maximum height of h and then starts falling down and finally touches the ground at point R units away from the throwing point.

Figure 1. The motion of an object thrown up from the ground at an angle.

The velocity of the particle t seconds after its projection is labeled as v_t. It is directed at \theta_t degrees from the horizon. It has v_t \cos\theta_t and v_t\sin\theta_t as the horizontal and vertical components.

Let us say at t = 0, v_0 = u. Further, let the initial projection direction to the ground (horizon) be represented as \theta_0 for simplicity. We label it as \theta. We divide the initial velocity into horizontal and vertical components, u\cos\theta and u\sin\theta. The projected object experiences only one force on it, the gravitational pull downwards, which produces acceleration g that is acting against the vertical component of projection velocity and hence decelerates the object until its vertical velocity becomes 0. At that point, the object attains height h above the horizon and starts falling. However, all the while, the horizontal component of the velocity finds no resistance (as gravity is acting at a right angle to it, we have ignored any effects of air resistance in our calculation). Let T_1 be the time the object takes to reach the maximum height h.

Following the formula [A], Final \hspace {3} Velocity = Initial \hspace {3}Velocity - Acceleration \times Time.

0= u\hspace {2}\sin\theta - g\times \hspace {3} T_1

\therefore T_1 = \frac {u\hspace {2}\sin\theta}{g} \hspace {60}-----[1]

Using formula [C], we write,

0^2 = (u\hspace {2}\sin\theta)^2 - 2 \times g \times h

hence, h = \frac{(u\hspace {2}\sin\theta)^2}{2g}  \hspace {30}-----[2]

Nature of projectile’s flight path

This was the calculation for the part the object was traveling up. For the descent, the initial velocity’s vertical component is zero, there is acceleration g which is now bringing it down, the horizontal component is undisturbed with value u\cos\theta and another known quantity is the distance to be traveled before the object touches ground h. Let T_2 be the time the object takes to fall from the top point while traversing the distance to the ground h. Now, using the equation [B],

h=0\times T_2 + \frac{1}{2}gT_2^2 = \frac{1}{2}gT_2^2 \hspace{7} -----[3]

Equating the value of h from [2] and [3],

\frac{(u\hspace {2}\sin\theta)^2}{2g} = \frac{1}{2}gT_2^2

\therefore T_2 = \frac {u\hspace {2}\sin\theta}{g}, which is the same as T_1 (confer with [1]).

We have, therefore, proved that the flight path or trajectory is symmetrical in time on both the sides of maximum height. This means the trajectory is a curve which is symmetrical on both the sides of the vertical line of height h. We can therefore write the equation for total time T of the flight,

T = T_1 + T_2 =   \frac {2 \hspace {2} u\sin\theta}{g} \hspace {30} -----[4]

The horizontal range of the projectile R is the horizontal velocity multiplied by total time T.

R = u\cos\theta \times \frac {2 \hspace {2} u\sin\theta}{g}

R = \frac {u^2} {g} \times (2 \hspace {2}\sin\theta \hspace {2} \cos\theta)

R = \frac {u^2 \sin 2\theta}{g} \hspace {75} -----[5]

We know that \sin 90^{\circ} = 1;  from relation [5] above we conclude that the maximum range would be attained by the projectile if it is thrown at an angle of 45^{\circ} from the horizon.

>Copying formula [3] here  and putting the value of T from  [4],

h= = \frac{1}{2}g(\frac {u\sin\theta}{g})^2

\therefore h= \frac {u^2{(\sin\theta})^2}{2g}              \hspace{60} -----[6]

What is the position of the object at time t after the launch? It is equal to the the velocity multiplied by time in case of uniform speed and in case of acceleration it is given by equation of motion [B].

x = u \times \cos \theta \times t  \hspace{0.75 in}    -----[7]

or, t = \frac {x} {u \cos \theta}  \hspace{1 in}     -----[7A]

y = u \sin\theta \times t - \frac{1}{2} \times g \times t^2 \hspace{10}   -----[8]

Upon putting the value of t from [7A] into [8] we get,

y = x  \tan \theta - \frac{1}{2}g \left( \frac{x}{u {\cos\theta}} \right)^2

y = \tan \theta \times x - \frac{g} {2 {(u \cos \theta)^2}} \times x^2 \hspace {1}  -----[9]

which is of the form,

y = Ax^2 + Bx + C    \hspace {40} -----[10]

– the standard equation of a parabola.

Let us give \theta some numerical value, say, 45^{\circ}, then in [8] we have \tan 45^{\circ} = 1, and \cos 45^{\circ} = \sin 45^{\circ} = \frac {1}{\sqrt 2} and for simplicity’s sake, we further assume g=1 (instead of 0.98) and u = 1 m/s.   The equation [9] becomes,

y = x - x^2

When we trace it, it clearly shows that it is a parabola. Please give some other values to \theta and experiment.

[Figure 2: Plotted using Mathpower]

However, a projectile need not be thrown up from and land on the ground (the zeroth item in the list below). There can be other situations.

Different situations

0. An object is thrown up from the ground at an angle.
1. An object is dropped from a height.
2. An object is thrown straight overhead.
3. An object is thrown horizontally from a height.
4. An object is thrown at such a speed and angle that it starts circumnavigating the Earth.

An object is dropped from a height.

Figure 3. An object is dropped from a height. 

Here u = 0\theta = 0, and acceleration is g; from equation [B] we get,

S= h = 0 \times\ t + \frac{1} {2} g  t^2

So, the time t to fall to the ground is \sqrt \frac{2 h} {g}.

The speed with which the object hits the ground is given by the formula [C] and is v =\sqrt {2gh}.

What if the object is thrown vertically upwards with speed u and reaches height h?

Figure 4. An object is thrown straight overhead.

The object slows down when its speed v = 0, hence equation [C] becomes, 0^2 = u^2 - 2 gh, we get the maximum height reached, h=\frac{u^2}{2g} and the time taken in doing so from relation [A], 0 = u - gt (u and g are directed oppositely), we get t = \frac{u}{g} and the total time from throwing up to falling back to the launch height t = \frac{2u}{g}.