A projectile fired into the first quadrant from the origin of a coordinate system will pass through the point (x, y) at time t according to the relationship, cos theta 2x/2y + gt2
where theta = the angle of elevation of the launcher and g = the acceleration due to gravity An artilleryman is firing at an enemy bunker located 2450 feet up the side of a hill that is 6175 feet away. He fires a round, and exactly 2.27 seconds later he scores a direct hit.
(a) What angle of elevation did he use?
(b) If the angle of elevation is also given by sec θ = υ0t/x. where υ0 is the muzzle velocity of the weapon, find the muzzle velocity of the artillery piece he used.

a. 22.3°

b. 2940.16 ft/s

Step-by-step explanation:

Here is the complete question

A projectile fired into the first quadrant from the origin of a coordinate system will pass through the point (x, y) at the relationship cot theta = 2x/2y + gt^2 where theta is the angle of elevation of the launcher and g is the acceleration due to gravity (32.2 feet/second^2). An artillery man is firing at an enemy barker located 2275 feet up the side of a hill that is 6325 feet away. He fires a round, and exactly 2.84 seconds later he scores a direct hit.

a. What angle of elevation did he use? theta = degree (Do not round until the final answer. Then round to one decimal place as needed.)

b. f the angle of elevation is also given by sec theta = v_0 t/x, where v_0 is the velocity of the weapon, find the muzzle velocity of the artillery piece he used. v_0 = feet/second (Round to two decimal places as needed.)

Solution

(a) What angle of elevation did he use?

Given that cosθ = 2x/(2y + gt²) where x = horizontal distance of projectile = 6175 feet away,y = vertical distance of projectile = 2450 feet up, g = acceleration due to gravity = 32 ft/s² and t = time of hit = 2.27 s

So, cosθ = 2x/(2y + gt²)

cotθ = 2 × 6175 ft/(2 × 2450 ft + 32 ft/s² × (2.27 s)²)

cotθ = 12350 ft/(4900 ft + 32 ft/s² × 5.1529 s²)

cotθ = 12350 ft/(4900 ft + 164.8928 ft)

cotθ = 12350 ft/(5064.8928 ft)

1/tanθ = 2.4384

tanθ = 1/2.4384

tanθ = 0.4101

θ = tan⁻¹(0.4101)

θ = 22.3°

(b) If the angle of elevation is also given by sec θ = υ0t/x. where υ0 is the muzzle velocity of the weapon, find the muzzle velocity of the artillery piece he used.

Making u0 subject of the formula, we have

u0 = xsecθ/t

u0 = x/tcosθ

since x = 6175ft, t = 2.27 s and θ = 22.3°, then

u0 = x/tcosθ

u0 = 6175 ft/(2.27 s × cos22.3°)

u0 = 6175 ft/(2.27 s × 0.9252)

u0 = 6175 ft/(2.1 s)

u0 = 2940.16 ft/s

d

step-by-step explanation:

its a straight line parallel to the time axis