Ask Question
30 November, 15:30

a projectile is thrown upward so that it's distance above the ground after T seconds is H equals - 16t^2 + 672 T. After how many seconds does it reach its maximum height?

+5
Answers (1)
  1. 30 November, 15:45
    0
    Step-by-step explanation:

    T = time in seconds

    H = distance

    Tground = time to return to ground

    Tmax = time at maximum height

    H = - 16T^2 + 672T ... eqn 1

    projectile returns to ground at H = 0,

    subs for H in eqn 1 ...

    0 = - 16T^2 + 672T

    solving for T we get ...

    16T^2 = 672T

    => Tground = 42secs

    Tmax = 0.5 Tground = 21secs
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “a projectile is thrown upward so that it's distance above the ground after T seconds is H equals - 16t^2 + 672 T. After how many seconds ...” in 📗 Mathematics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers