Question

A hunter aims at a deer which is 40 yards away. Her cross- bow is at a height of 5ft, and she aims for a spot on the deer 4ft above the ground. The crossbow fires her arrows at 300ft/s. (a) At what angle of elevation should she hold the cross- bow to hit her target? (b) If the deer is moving perpendicularly to her line of sight at a rate of 20mph, by approximately how much should she lead the deer in order to hit it in the de- sired location?

Answer 1

a) θ₁ = 0.487º , b) t = 0.400 s , x = 11.73 ft

Step-by-step explanation:

For this exercise let's use the projectile launch relationships.

The initial height is I = 5 ft and the final height y = 4 ft

y = y₀ +
`v_(oy)` t - ½ g t²

The distance to the band is x = 40 yard (3 ft / 1 yard) = 120 ft

x = v₀ₓ t

t = x / v₀ₓ

We replace

y –y₀ = v_{oy} x / v₀ₓ - ½ g x² / v₀ₓ²

v_{oy} = v₀ sin θ

v₀ₓ = vo cos θ

y –y₀ = x tan θ - ½ g x² / v₀² cos² θ

5-4 = 120 tan θ - ½ 32 120 / (300 2 cos2 θ)

1 = 120 tan θ - 0.0213 sec² θ

Let's use the trigonometry relationship

Sec² θ = 1 - tan² θ

1 = 120 tan θ - 0.0213 (1 –tan²θ)

0.0213 tan²θ + 120 tanθ -1.0213 = 0

We change variables

u = tan θ

u² + 5633.8 u - 48.03 = 0

We solve the second degree equation

u = [-5633.8 ±√(5633.8 2 + 4 48.03)] / 2

u = [- 5633.8 ± 5633.82] / 2

u₁ = 0.0085

u₂= -5633.81

u = tan θ

θ = tan⁻¹ u

For u₁

θ₁ = tan⁻¹ 0.0085

θ₁ = 0.487º

For u₂

θ₂ = -89.99º

The launch angle must be 0.487º

b) let's look for the time it takes for the arrow to arrive

x = v₀ₓ t

t = x / v₀ cos θ

t = 120 / (300 cos 0.487)

t = 0.400 s

The deer must be at a distance of

v = 20 mph (5280 ft / 1 mi) (1 h / 3600s) = 29.33 ft / s

x = v t

x = 29.33 0.4

x = 11.73 ft