SOLUTION: h(t)=c-(d-4t)(d-4t)
At time t=0, a ball was thrown upward from an initial height of 6 feet. until the ball hit the ground, its height, in feet, after t seconds was given by the f
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-> SOLUTION: h(t)=c-(d-4t)(d-4t)
At time t=0, a ball was thrown upward from an initial height of 6 feet. until the ball hit the ground, its height, in feet, after t seconds was given by the f
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Question 296264: h(t)=c-(d-4t)(d-4t)
At time t=0, a ball was thrown upward from an initial height of 6 feet. until the ball hit the ground, its height, in feet, after t seconds was given by the function h above, in which c and d are positive constants. if the ball reached its maximum height, of 106 ft at time t+2.5, what was the height, in feet, of the ball at time t=1. Answer by richwmiller(17219) (Show Source):
You can put this solution on YOUR website! h(t)=c-(d-4t)(d-4t)
t=2.5
h(2.5)=106
h(0)=6
6=c-(d)^2
106=c-(d-10)^2
c=106 d=10
h(t)=106-(10-4t)^2
h(1)=70
