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

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