SOLUTION: 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) = 
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Question 1007222: 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) = −16t^2 + v0t + s0.
If s(2) = 131 and s(4) = 89, what are v0 and s0?
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
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) = −16t^2 + v0t + s0.
If s(2) = 131 and s(4) = 89, what are v0 and s0?
:
Write an equation for each pair of values
t=2; s=131
-16(2^2) + 2Vo + So = 131
-64 + 2Vo + So = 131
2Vo + So = 131 + 64
2Vo + So = 195
and
t=4; s=89
-16(4^2) + 4Vo + So = 89
-256 + 4Vo + So = 89
4Vo + So = 89 + 256
4Vo + So = 345
:
Use elimination on these two equations
4Vo + So = 345
2Vo + So = 195
-----------------Subtraction eliminated So, find Vo
2Vo = 150
Vo = 75 ft/sec is the upward velocity
:
Find So
2(75) + So = 195
150 + So = 195
So = 195- 150
So = 45 ft is the initial height above the ground
:
The equation
S(t) = -16t^2 + 75t + 45
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