Chapter 3, Problem 21P

The maximum distance s and the maximum height h that a projectile shot at an angle θ are given by:

$s=\frac{v^{2}0}{g}\sin$ 20 and $h=\frac{v_0^2{\sin }^2\theta }{2g}$

where is the shooting velocity and g=9,81 m/s². Determine s(θ) and h(θ) for

θ = 15°, 25°, 35° 45°, 55°.65°, 75° if = 260m/s.