Question

When a mortar shell is fired with an initial

velocity of v0 ft/sec at an angle α above the
horizontal, then its position after t seconds is
given by the parametric equations
x = (v0 cos α)t , y = (v0 sin α)t − 16t^2.
If the mortar shell hits the ground 2500 feet
from the mortar when α = 75degress, determine v0
answer choices
1. v0 = 360 ft/sec
2. v0 = 370 ft/sec
3. v0 = 380 ft/sec
4. v0 = 390 ft/sec
5. v0 = 400 ft/sec

Answer 1

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.

Answer 2

The equation y = Vo cos(alfa) t - 16t^2 implies that the mortar landed at the same level that it was fire and that fire angle is also 75°.

With that said, let us work on the parametric equations

y = 0 = Vo sin(alfa) t - 16t^2, which by factoring =>

t (Vo sin(alfa) - 16t) = 0 => t = Vo sin(alfa) / 16 .....(1)

x = 2500 = Vo cos(alfa) t => t = 2500 / [Vo cos(alfa)] ..... (2)

Now make (1) equal to (2)

Vo sin(alfa) / 16 = 2500 / [Vo cos(alfa)] =>

Vo^2 sin(alfa)cos(alfa) = 2500ft*16ft/s^2 =>

Vo^2 = 2500*16 / [sin(75°)cos(75)] = 2500*16/0.25 = 160,000 ft^2/s^2

Vo = √(160,000) ft/s = 400 ft/s

Answer: option 5. 400 ft/s