# 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**

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 .

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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