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20 March, 05:54

If a rock is thrown upward on the planet Mars with a velocity of 18 m/s, its height (in meters) after t seconds is given by H = 18t - 1.86t2. (a) Find the velocity of the rock after one second. 14.28 Correct: Your answer is correct. m/s (b) Find the velocity of the rock when t = a. Correct: Your answer is correct. m/s (c) When will the rock hit the surface? (Round your answer to one decimal place.) t = Incorrect: Your answer is incorrect. s (d) With what velocity will the rock hit the surface? m/s

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  1. 20 March, 06:01
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    a) The velocity of the rock after 1 s is 14.28 m/s.

    b) The velocity of the rock after "a" seconds is 18 m/s - 3.72 m/s² · a

    c) The rock will hit the ground after 9.7 s.

    d) The velocity with which the rock hits the surface is - 18.1 m/s

    Explanation:

    Hi there!

    Let's write the equation:

    H (t) = 18 m/s · t - 1.86 m/s² · t²

    The velocity is the variaiton of the height over time, in other words, it is the derivative of the height function with respect to time. Then:

    v (t) = dH/dt = 18 m/s - 2 · 1.86 m/s² · t

    v (t) = 18 m/s - 3.72 m/s² · t

    a) When t = 1 s:

    v (1 s) = 18 m/s - 3.72 m/s² · 1 s

    v (1 s) = 14.28 m/s

    b) When t = a:

    v (a) = 18 m/s - 3.72 m/s² · a

    c) When the rock hits the surface, the height of the rock is 0:

    H (t) = 18 m/s · t - 1.86 m/s² · t²

    0 = 18 m/s · t - 1.86 m/s² · t²

    0 = t (18 m/s - 1.86 m/s² · t)

    t = 0

    and

    18 m/s - 1.86 m/s² · t = 0

    t = - 18 m/s / - 1.86 m/s²

    t = 9.7 s

    d) Let's use the equation of velocity to find the velocity at t = 9.7 s when the rock hits the ground.

    v (t) = 18 m/s - 3.72 m/s² · t

    v (9.7 s) = 18 m/s - 3.72 m/s² · 9.7 s

    v (9.7 s) = - 18.1 m/s
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