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# The maximum range of projectile formula

Let us consider a projectile projected with initial velocity $$v_{0}$$ making an angle $$\theta_0$$ with the horizontal as shown below in the figure.

We know that the horizontal range of a projectile is the distance traveled by the projectile during its time of flight. This horizontal range is given by the relation $\text{Horizontal Range}=\text{Horizontal velocity}\times \text{time of flight}$ So, the formula for the horizontal range is $R=\frac{{v_0}^2 \sin 2\theta_0}{g} \qquad (1)$

## The maximum range for projectile motion

A projectile of the same mass can be launched with the same initial velocity and different angles $$\theta_0$$.
Consider the figure given below where a projectile is launched with three different angles $$45^0,\,\,60^0,\,\,and\,\,30^0$$. A projectile launched at different angles but the same initial velocity

From the above figure, we can clearly see that for different angles of projection horizontal range of a projectile is different.
Let us now investigate the angle of projection for which range is maximum (note that initial velocity remains the same).

$R=\frac{v_0 \sin 2\theta_0}{g}$ In above relation, if we keep initial velocity same for various launch , then value of $$R$$ varies with the value of $$\sin 2\theta_0$$.
Clearly, from trigonometry, we know that the maximum value of $$\sin \theta=1$$ and we know that $$\sin 90^0=1$$
So in this case, we have
$$2\theta_0=90^0 \qquad or,\qquad \theta_0=45^0$$
Thus the horizontal range of a projectile for a given initial velocity is maximum when it is projected at an angle of $$45^0$$ with the horizontal.

### The maximum range of projectile formula

To find the formula for a maximum range put $$\theta_0=45^0$$ in equation (1). So, from equation (1) we have
$R_m=\frac{v_0^2 \sin 90^0}{g}=\frac{v_0^2}{g}$

#### Question That can be asked from this topic

Question 1: Why do 45 degrees give a maximum range of a projectile?
Question 2: Prove 45 degrees maximum range for a projectile launched at an angle $$\theta$$.
Question 3: Derive the formula for the range of projectile body launched at an angle $$\theta$$.