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

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