+2498 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 ?
+128460 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
