Question

A projectile's horizontal range on level ground is r=v20sin2θ/g. at what launch angle or angles will the projectile land at half of its maximum possible range. enter your answers numerically separated by commas. express your answer using two significant figures.

Answer 1

As given the formula of range is

$R = \frac{v^2sin2\theta}{g}$

now for the maximum range

$\theta = 45$

$R_{max} = \frac{v^2}{g}$

now for half of this maximum range we will have

$\frac{R_{max}}{2} = \frac{v^2sin2\theta}{g}$

$\frac{v^2}{2g} = \frac{v^2sin2\theta}{g}$

now from above we have

$sin2\theta = \frac{1}{2}$

so two possible values for above is given as

$\theta = 15 degree, 75 degree$

so above two angle we will have half of maximum range