10
advanced-math

A ball is thrown with height h(t)=16t2+32t+10h(t) = -16t^2 + 32t + 10. When does it hit the ground (h=0h=0)?

A

2.28 seconds

B

1.00 seconds

C

0.27 seconds

D

4.50 seconds

Correct Answer: A

Choice A is the correct answer. We need to solve 16t2+32t+10=0-16t^2 + 32t + 10 = 0.

  1. Discriminant: 3224(16)(10)=1024+640=166432^2 - 4(-16)(10) = 1024 + 640 = 1664.
  2. Formula: t=32±166432t = \frac{-32 \pm \sqrt{1664}}{-32}.
  3. Approximate 166440.8\sqrt{1664} \approx 40.8.
  4. t32±40.832t \approx \frac{-32 \pm 40.8}{-32}.
    • Case 1: (72.8)/322.275(-72.8)/-32 \approx 2.275
    • Case 2: (8.8)/320.275(8.8)/-32 \approx -0.275 (Time cannot be negative)

So, t2.28t \approx 2.28 seconds.

Choice B is incorrect because t=1t=1 is the time of maximum height (vertex). Choice C is incorrect because it corresponds to the negative root magnitude. Choice D is incorrect because it is too large.