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# one mistake and bah ! :)

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A physics student charts the path of a ball thrown into the air. He finds that the vertical height, V(t), of the ball at time t can be modeled using the following quadratic equation: V(t)=I-(t-h)^2, where I and h are unknown constants. The initial heigt of the ball at time t=0 is 9 ft. and the ball reaches its maximum height of 25 ft. at time t=4. What is the height of the ball at time t=2 ?

Dec 18, 2015

### 3+0 Answers

#1
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V(t)=I-(t-h)^2    at t =0 ,  V(0)  = 9     so we have

9 =  l - (0 -h)^2

9 = l  - h^2      rearranging.........l = 9 + h^2

At t = 4, V(4)  = 25     and we have

25 = [9 + h^2] - ( 4 - h)^2   simplify

25 = 9 + h^2 - h^2 + 8h - 16

25 = 8h -7

32  = 8h

h = 4         so    l  =[ 9 + 4^2] = [9 + 16]  =  25

And the height at t = 2   is given by

V(2)  = 25 - (2 - 4)^2  =    25 - (-2)^2   =    25 - 4   =    21   ft

Here's the graph.......https://www.desmos.com/calculator/zucm57abuz   Dec 18, 2015
#2
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pfffff i forget the - :(

Thanks CPhill

Dec 18, 2015
#3
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No prob.......   Dec 18, 2015