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 $45^0,\,\,60^0,\,\,and\,\,30^0$.

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).

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$$R=\frac{v_0 \sin 2\theta_0}{g}$$ 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 $\theta$.