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.

Projectile Motion |

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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

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 |

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\[R_m=\frac{v_0^2 \sin 90^0}{g}=\frac{v_0^2}{g}\]

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