Ask Question
24 January, 02:08

If an arrow is shot upward on the moon with a velocity of 58 m/s, its height (in meters) after t seconds is given by H = 58t - 0.83t2. (a) Find the velocity of the arrow after two seconds. m/s (b) Find the velocity of the arrow when t = a. m/s (c) When will the arrow hit the surface? (Round your answer to one decimal place.) t = s (d) With what velocity will the arrow hit the surface? m/s

+2
Answers (1)
  1. 24 January, 05:02
    0
    a) v = 54.7m/s

    b) v = (58 - 1.66a) m/s

    c) t = 69.9 s

    d) v = - 58.0 m/s

    Explanation:

    Given;

    The height equation of the arrow;

    H = 58t - 0.83t^2

    (a) Find the velocity of the arrow after two seconds. m/s;

    The velocity of the arrow v can be given as dH/dt, the change in height per unit time.

    v = dH/dt = 58 - 2 (0.83t) ... 1

    At t = 2 seconds

    v = dH/dt = 58 - 2 (0.83*2)

    v = 54.7m/s

    (b) Find the velocity of the arrow when t = a. m/s

    Substituting t = a into equation 1

    v = 58 - 2 (0.83*a)

    v = (58 - 1.66a) m/s

    (c) When will the arrow hit the surface? (Round your answer to one decimal place.) t = s

    the time when H = 0

    Substituting H = 0, we have;

    H = 58t - 0.83t^2 = 0

    0.83t^2 = 58t

    0.83t = 58

    t = 58/0.83

    t = 69.9 s

    (d) With what velocity will the arrow hit the surface? m/s

    from equation 1;

    v = dH/dt = 58 - 2 (0.83t)

    Substituting t = 69.9s

    v = 58 - 2 (0.83*69.9)

    v = - 58.0 m/s
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “If an arrow is shot upward on the moon with a velocity of 58 m/s, its height (in meters) after t seconds is given by H = 58t - 0.83t2. (a) ...” in 📙 Physics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers