Final answer:
To determine the initial velocity, v0, we use the given position equations and the fact that the mortar shell hits the ground 2500 feet from the mortar when α = 75 degrees. Solving for v0 gives an approximate value of 370 ft/sec.
Step-by-step explanation:
To determine the initial velocity, v0, we can use the given information that the mortar shell hits the ground 2500 feet from the mortar when α = 75 degrees. Using the parametric equations for the position of the mortar shell, we can substitute the given values: x = (v0 cos α)t
y = (v0 sin α)t - 16t^2
Since the mortar shell hits the ground, the y-coordinate is 0. Plugging this into the second equation, we get: 0 = (v0 sin α)t - 16t^2
Using the angle α = 75 degrees and solving for t gives us: t = (v0 sin α) / 16
Substituting this value of t into the x-coordinate equation, we get:
2500 = (v0 cos α) * (v0 sin α) / 16
Multiplying both sides by 16 and using the identity sin α cos α = sin(2α)/2, we can solve for v0:
40000 = v0^2 * sin(2α)/2
v0^2 = 80000 / sin(2α)
v0 = sqrt(80000 / sin(2α))
Plugging in α = 75 degrees, we can calculate:v0 = sqrt(80000 / sin(150)) ≈ 370 ft/sec
Therefore, the correct answer is v0 = 370 ft/sec.