Projectile Motion

Question

The speed of an arrow fired from a compound bow is about 88 m/s. An archer sits astride his horse and launches an arrow into the air, elevating the bow at an angle of 38◦ above the horizontal and 1.5 m above the ground. What is the arrow’s range? The acceleration of gravity is 9.81 m/s 2 . Assume the ground is level and ignore air resistance. Answer in units of m.

Answer 1

The equation of motion of the arrow's location (x,y) is d²(x,y)/dt² = (0,–g) due to constant gravitational acceleration. Its first integral gives the velocity vector as d(x,y)/dt = (cosθ,sinθ)s + (0,–g)t, and its second integral gives the location vector as (x,y) = (0,y₀) + (cosθ,sinθ)st + (0,–g)t²/2. Therefore the time t → T when y → 0 will be the positive root of 0 = y₀ + st sinθ – gt²/2 = –(t² – 2st/g sinθ – 2y₀/g)g/2 = –((t – s/g sinθ)² – 2y₀/g – s²/g² sin²θ)g/2, that is T = s/g sinθ + √(2y₀/g + s²/g² sin²θ). At t = T, x is at the arrow's range.