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?
+130466 We know that, at t = 0, we have
6 = c - ( d - 4(0))^2
6 = c -(d)^2
So...... c = 6 + d^2
And we also know that, at t = 2.5, we have
106 = c - (d -4(2.5) )^2 simplify
106 = c - (d - 10)^2
106 = c - d^2 + 20d - 100 and substituting for c, we have
106 = 6 + d^2 - d^2 + 20d - 100
100 = 20d - 100
200 = 20d → so d = 10 and c = 6 + 10^2 = 106
So.....at t = 1 we have
h(1) = 106 - (10 - 4(1) )^2
h(1) = 106 - (6)^2
h(1) = 106 - 36 = 70 ft. when t = 1
Here's a graph : https://www.desmos.com/calculator/f1v94eldv1
