# SOLUTION: h(t)= -16t^2+dt+c At time t=0, a rocket was launched upward from an initial height of 6 feet. Until the rocket hit the ground, its height, in feet, after t seconds was gi

Algebra ->  Algebra  -> Expressions-with-variables -> SOLUTION: h(t)= -16t^2+dt+c At time t=0, a rocket was launched upward from an initial height of 6 feet. Until the rocket hit the ground, its height, in feet, after t seconds was gi      Log On

 Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations! Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help!

 Algebra: Expressions involving variables, substitution Solvers Lessons Answers archive Quiz In Depth

 Click here to see ALL problems on Expressions-with-variables Question 630072: h(t)= -16t^2+dt+c At time t=0, a rocket was launched upward from an initial height of 6 feet. Until the rocket hit the ground, its height, in feet, after t seconds was given by the function h aobve, in which c and d are positive constants. If the rocket reached its maximum height of 350 ft at time t=4, what was the height, in feet, of the rocket at time t=1? (the answer is 140, by the way) please include explanations!Answer by stanbon(57262)   (Show Source): You can put this solution on YOUR website!h(t)= -16t^2+dt+c At time t=0, a rocket was launched upward from an initial height of 6 feet. Until the rocket hit the ground, its height, in feet, after t seconds was given by the function h aobve, in which c and d are positive constants. If the rocket reached its maximum height of 350 ft at time t=4, what was the height, in feet, of the rocket at time t=1? ------ h(0) = -16*0^2 + d*0 + c = 6 feet So, c = 6 --------------------- h(4) = 350 h(4) = -16*4^2+4d + 6 = 350 256 + 4d + 6 = 350 4d = 88 d = 22 ------ Equation: h(t) = -16t^2 + 22t + 6 h(1) = -16 + 22 + 6 h(1) = 12 feet ===================== Cheers, Stan H.