Projectile fired at an angle with Horizontal

If a body fall freely under the influence of gravity, then it is called projectile. This motion is called projectile motion.

Projectile fired at an angle with horizontal

Projectile motion

Let us consider an object is projected with the initial velocity u at an angle θ with the ground taken as x-axis. OY is a vertical line perpendicular to the ground. The velocity of an object u has two components ucosθ along x axis and usinθ along y axis. So initial velocity in horizontal direction is ucos θ and vertically upward direction is usinθ .

The motion of projectile is two dimensional. So, the horizontal distance is covered by horizontal velocity and the vertical distance is due to the vertical velocity.

Note: Vertical velocity is affected by gravity while horizontal velocity is not affected by gravity.
Now, Let the object is at a point P in time t whose horizontal and vertical distances are x and y. As horizontal distance is constant, horizontal distance X in time t is given by:
$$x = ucos \theta \times t$$
$$or, t = \frac{x}{ucos \theta }…..(i)$$
Since, vertical velocity is affected by gravity, the vertical distance y covered in time t is given as:
$$y = usin \theta t – \frac{1}{2} gt^2$$
Putting the value of t from equation (i), we get:
$$y = usin \theta \times \frac{x}{ucos \theta } – \frac{1}{2} g(\frac{x}{ucos \theta })^2$$
$$or, y = xtan \theta – \frac{g}{2u^2cos^2 \theta } \times x^2 $$
$$or, y = px-qx^2……(ii)$$
$$ \text{Where, p = tan} \theta$$
$$ q = \frac{g}{2u^2cos^2 \theta }$$
Equation (ii) is the equation of parabola. Hence, it is proved that the path of projectile is parabolic.


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