If an object is projected vertically upward from an altitude of s0 feet with an initial velocity of v0 ft/sec?

Question

If an object is projected vertically upward from an altitude of s0 feet with an initial velocity of v0 ft/sec, then its distance s(t) above the ground after t seconds is
s(t) = −16t2 + v0t + s0.
If s(1) = 94 and s(4) = 109, what are v0 and s0?
v0 =

s0 =

Answer 1

s(t) = −16t2 + v0t + s0.

If s(1) = 94 and s(4) = 109, what are v0 and s0?







Substiute 1 for t and 94 for s(t)



94 = -16(1^2) + v0 (1) +s0



substitue 4 for t and 109 for for s(t)



109 = -16 (4^2) + v0 (4) +s0



you now have two equations in two unknowns:



eq1) 94 = -16(1^2) + v0 (1) +s0

eq2) 109 = -16 (4^2) + v0 (4) +s0





to find v0 subtract eq1 from eq2 to eliminate s0



109 - 94 = -16 (4^2)- (-16(1^2))+ 4v0 -v0 + s0-s0

15 = -240 + 3v0



v0 = 85



use V0 = 85 in either eq1 or eq2 to find s0

Answer 2

s(t) =--16t^2+v0t+s0

at t=1 sec s(1)=94 ft

therefore 94= -16*1^2+v0*1+s0 => v0+s0=110......(1)

at t=4 sec s(4)=109 ft

therefore 109= -16*4^2+4V0+s0 =>4v0+s0=365.....(2)

(2)-(1)

3v0=255 => v0=85 ft/sec

putting v0 in (1)

s0=110-85=25 ft

Answer 3

s(t) = - 16t² + v0t + s0 s(a million) = ninety 4: ninety 4 = - sixteen(a million)² + v0(a million) + s0 ninety 4 = - sixteen + v0 + s0 v0 + s0 = ninety 4 + sixteen v0 + s0 = one hundred ten s0 = one hundred ten - v0 ................. Eq. a million s(4) = 109: 109 = - sixteen(4)² + v0(4) + s0 109 = - sixteen(sixteen) + 4v0 + s0 109 = - 256 + 4v0 + s0 4v0 + s0 = 109 + 256 4v0 + s0 = 365 ............... Eq. 2 Sub one hundred ten - v0 from Eq. a million for s0 in Eq. 2: 4v0 + one hundred ten - v0 = 365 4v0 - v0 = 365 - one hundred ten 3v0 = 255 v0 = 255 / 3 v0 = 80 5 ¯¯¯¯¯¯ Sub 80 5 for v0 in Eq. a million: s0 = one hundred ten - 80 5 s0 = 25 ¯¯¯¯¯¯¯ Equation: s(t) = - 16t² + 85t + 25 ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯