Ask Question
2 November, 16:53

H (t) = - 5t^2+20t+1

what time does the ball reach the same height it was kicked at again? When does the ball reach its max height? What is the max height?

+3
Answers (1)
  1. 2 November, 18:22
    0
    H (t) = - 5t² + 20t + 1

    -5t² + 20t + 1 = 0

    t = - (20) + / - √ ((20) ² - 4 (-5) (1))

    2 (-5)

    t = - 20 + / - √ (400 + 20)

    -10

    t = - 20 + / - √ (420)

    -10

    t = - 20 + / - 2√ (105)

    -10

    t = - 20 + 2√ (105) t = - 20 - 2√ (105)

    -10 - 10

    t = 2 - 0.2√ (105) t = 2 + 0.2√ (105)

    h (t) = - 5t² + 20t + 1

    h (2 - 0.2√ (105)) = - 5 (2 - 0.2√ (105)) ² + 20 (2 - 0.2√ (105)) + 1

    h (2 - 0.2√ (105)) = - 5 (2 - 0.2√ (105)) (2 - 0.2√ (105)) + 20 (2) - 20 (0.2√ (105)) + 1

    h (2 - 0.2√ (105)) = - 5 (4 - 0.4√ (105) - 0.4√ (105) + 0.04√ (11025)) + 40 - 4√ (105) + 1

    h (2 - 0.2√ (105)) = - 5 (4 - 0.8√ (105) + 0.04 (105)) + 40 + 1 - 4√ (105)

    h (2 - 0.2√ (105)) = - 5 (4 - 0.8√ (105) + 4.2) + 41 - 4√ (105)

    h (2 - 0.2√ (105)) = - 5 (4 + 4.2 - 0.8√ (105)) + 41 - 4√ (105)

    h (2 - 0.2√ (105)) = - 5 (8.2 - 0.8√ (105)) + 41 - 4√ (105)

    h (2 - 0.2√ (105)) = - 5 (8.2) - 5 (-0.8√ (105)) + 41 - 4√ (105)

    h (2 - 0.2√ (105)) = - 41 + 4√ (105) + 41 - 4√ (105)

    h (2 - 0.2√ (105)) = - 41 + 41 + 4√ (105) - 4√ (105)

    h (2 - 0.2√ (105)) = 0 + 0

    h (2 - 0.2√ (105)) = 0

    (t, h (t)) = (2 - 0.2√ (105), 0)

    or

    h (t) = - 5t² + 20t + 1

    h (2 + 0.2√ (105)) = - 5 (2 + 0.2√ (105)) ² + 20 (2 + 0.2√ (105)) + 1

    h (2 + 0.2√ (105)) = - 5 (2 + 0.2√ (105)) (2 + 0.2√ (105)) + 20 (2) + 20 (0.2√ (105)) + 1

    h (2 + 0.2√ (105)) = - 5 (4 + 0.4√ (105) + 0.4√ (105) + 0.04√ (11025)) + 40 + 4√ (105) + 1

    h (2 + 0.2√ (105)) = - 5 (4 + 0.8√ (105) + 0.04 (105)) + 40 + 4√ (105) + 1

    h (2 + 0.2√ (105)) = - 5 (4 + 0.8√ (105) + 4.2) + 40 + 1 + 4√ (105)

    h (2 + 0.2√ (105)) = - 5 (4 + 4.2 + 0.8√ (105)) + 41 + 4√ (105)

    h (2 + 0.2√ (105)) = - 5 (8.2 + 0.8√ (105)) + 41 + 4√ (105)

    h (2 + 0.2√ (105)) = - 5 (8.2) - 5 (0.8√ (105)) + 41 + 4√ (105)

    h (2 + 0.2√ (105)) = - 41 - 4√ (105) + 41 + 4√ (105)

    h (2 + 0.2√ (105)) = - 41 + 41 - 4√ (105) + 4√ (105)

    h (2 + 0.2√ (105)) = 0 + 0

    h (2 + 0.2√ (105)) = 0

    (t, h (t)) = (2 + 0.2√ (105), 0)
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “H (t) = - 5t^2+20t+1 what time does the ball reach the same height it was kicked at again? When does the ball reach its max height? What is ...” in 📘 Mathematics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers