Answer to Question #294888 in Calculus for chan

Question

When the air resistance is ignored, the horizontal range R of a projectile is given by R(θ) = v2 0 g sin 2θ where v0 is the constant initial velocity, g is the acceleration due to gravity, and θ is the angle of elevation or departure. Find the maximum range of projectile.

Answer 1

Solution:

$R=\dfrac{v_{0}^{2} \sin 2 \theta}{g}$

This equation shows that for a given projectile velocity $\mathrm{v}_{0}, \mathrm{R}$  is maximum when $\sin 2 \theta$  is maximum, i.e. when $\theta_{0}=45^{\circ}$ . The maximum horizontal range is, therefore

${R}_{\mathrm{m}}=\dfrac{{v}_{0}{ }^{2}}{{~g}}$