SOLUTION: h(t)= c-(d-4t)^2
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 fun
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Question 389681: h(t)= c-(d-4t)^2
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 feet at time t= 2.5, what was the height, in feet, of the ball at time t = 1?
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
\ =\ c\ -\ \left(d\ -\ 4t\right)^2)
Expand the binomial:
\ =\ -16t^2\ +\ 8dt\ +\ c\ -\ d^2)
Compare this to the standard form of the height function for a projectile near the surface of the Earth with respect to time:
\ =\ -16t^2\ +\ v_ot\ +\ h_o)
By comparison you can see that the initial height, 
, must be equal to 
, and the initial velocity, 
.
We know that the maximum height is obtained at the time equal to the 
-coordinate of the vertex of the parabola that is the graph of the function. The -coordinate is found by:
but we were given that 
From which we can deduce:
(verification left as an exercise for the student)
and then we can determine that
Next we were given that 
, but knowing that 
, we can deduce that:
Plug in the values:
\ =\ 106\ -\ \left(10\ -\ 4(1)\right)^2)
And then do the arithmetic.
John
My calculator said it, I believe it, that settles it
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