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

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

experimentx · Answer

v^2 = u^2 sin^2(E) + 2 a s http://www.wolframalpha.com/input/?i=+0+%3D+x%5E2+-+2+*+9.8+*+762 u sin75 = 122.21 u = 126.5 m/s http://www.wolframalpha.com/input/?i=122.21%2Fsin%2875+deg%29 415.05 ft/s https://www.google.com/search?rlz=1C1ASUT_enNP447NP447&sourceid=chrome&ie=UTF-8&q=2500+ft+%3D%3F#hl=en&rlz=1C1ASUT_enNP447NP447&sa=X&psj=1&ei=BkeKT-GJJtHyrQfQ1qTUCw&ved=0CBcQBSgA&q=convert+126.5+meters+into+feet&spell=1&bav=on.2,or.r_gc.r_pw.r_cp.r_qf.,cf.osb&fp=7db0b50b83c4e821&biw=1366&bih=667

experimentx · Answer

There must be some error.

anonymous · Answer

exactly 400m/s

anonymous · Answer

$$u ^{2}=Rg/\sin2\alpha$$
solving for u gives u =$$\sqrt{160000}=400ft/s$$
remember g=32ft/s

anonymous · Answer

$$Range(R)=u ^{2}\sin2\alpha/g$$
here range=2500ft
$$\alpha$$=75

experimentx · Answer

must be right.