the equation h=−16t2+32t+9 gives the height of a ball, h, in feet above the ground, at t seconds after the ball is thrown upward. After how much time will it reach a height of 20 feet?

Answer 1

To find the time it takes for the ball to reach a height of 20 feet, we need to set the equation h = 20 and solve for t.

Substituting 20 for h in the equation h = -16t^2 + 32t + 9, we get:
20 = -16t^2 + 32t + 9

Rearranging the equation, we have:
16t^2 - 32t - 11 = 0

To solve this quadratic equation, we can use the quadratic formula: t = (-b ± √(b^2 - 4ac)) / (2a)
Where a = 16, b = -32, and c = -11.

Plugging in these values, we get:
t = (-(-32) ± √((-32)^2 - 4(16)(-11))) / (2(16))
t = (32 ± √(1024 + 704)) / 32
t = (32 ± √1728) / 32

Simplifying further, we have:
t = (32 ± 41.57) / 32

Taking both the positive and the negative square root values, we get two possible values for t:
t = (32 + 41.57) / 32 ≈ 2.30 seconds
t = (32 - 41.57) / 32 ≈ -0.29 seconds

Since time cannot be negative, we discard the negative solution.

Therefore, the ball will reach a height of 20 feet after approximately 2.30 seconds.