A basketball is thrown from a balcony. Its height in feet after t seconds is h(t) = -16t^2 + 32t + 6. a....

Question

7) A basketball is thrown from a balcony. Its height in feet after t seconds is h(t) = -16t^2 + 32t + 6.
a. How high is the balcony?
b. What is the maximum height, and when does it happen?
c. How long until the ball hits the ground?
d. How high is the ball after 0.5 seconds?

SnapAndSolve AI · Accepted Answer

a. The height of the balcony is the initial height when t = 0. Substitute t = 0 into the equation: h(0) = -16(0)^2 + 32(0) + 6 = 6 feet. The balcony is 6 feet high.
b. The maximum height occurs at the vertex of the parabola. The time at which this occurs is given by t = -b/(2a) where a = -16 and b = 32. So, t = -32/(2*(-16)) = 1 second. Substitute t = 1 into the equation to find the maximum height: h(1) = -16(1)^2 + 32(1) + 6 = 22 feet. The maximum height is 22 feet, and it happens at 1 second.
c. The ball hits the ground when h(t) = 0. Solve -16t^2 + 32t + 6 = 0 using the quadratic formula: t = [-b ± sqrt(b^2 - 4ac)] / 2a, where a = -16, b = 32, and c = 6. The solutions are t = 2.25 and t = -0.167. Since time cannot be negative, the ball hits the ground at t = 2.25 seconds.
d. Substitute t = 0.5 into the equation to find the height: h(0.5) = -16(0.5)^2 + 32(0.5) + 6 = -4 + 16 + 6 = 18 feet. The ball is 18 feet high after 0.5 seconds.

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A basketball is thrown from a balcony. Its height in feet after t seconds is h(t) = -16t^2 + 32t + 6. a....

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