WAEC Physics 2017 Paper

Concept:For projectile motion, the horizontal range is maximum when the launch angle makes the sine term in the range formula equal to its greatest possible value.Explanation:The range of a projectile is given by $R = \frac{v^{2}\sin 2\theta}{g}$.Here, $v$ is the initial speed, $\theta$ is the angle of projection, and $g$ is the acceleration due to gravity.For fixed $v$ and $g$, the value of $R$ is maximum when $\sin 2\theta$ is maximum.The maximum value of $\sin 2\theta$ is $1$, which occurs when $2\theta = 90^\circ$.Solving gives $\theta = \frac{90^\circ}{2} = 45^\circ$.Answer:$\theta = 45^\circ$, so the correct option is B.