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If a rock is thrown vertically upward from the surface of Mars with velocity of 20 m/s, its height (in meters) after t seconds is h = 20t − 1.86t2. (a) What is the velocity (in m/s) of the rock after 2 s? 12.56 Correct: Your answer is correct. m/s (b) What is the velocity (in m/s) of the rock when its height is 25 m on its way up? On its way down? (Round your answers to two decimal places.) up 8.07 Incorrect: Your answer is incorrect. m/s down -8.07 Incorrect: Your answer is incorrect. m/s

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onyebuchinnaji

The velocity of the rock after 2s is 12.56 m/s.

The velocity of the rock at 25 m upwards is 14.68 m/s.

The velocity of the rock at 25 m downwards is -15.97 m/s.

The given parameters;

  • velocity of the rock, u = 20 m/s
  • height of the rock, h = 20t - 1.8t²

The velocity of the rock after 2s is calculated as follows;

[tex]v = \frac{dh}{dt}= 20 -2(1.86)t \\\\v = 20 - 3.72t\\\\at \ t= 2 \ s \\\\v = 20 - 3.72(2) = 12.56 \ m/s[/tex]

The velocity of the rock when the height is 25 m:

h = 20t - 1.8t²

25 = 20t - 1.8t²

1.8t² - 20t + 25 = 0

solve for the time of motion "t" using quadratic formula

a = 1.8,  b = -20,  c = 25

[tex]t = \frac{-b \ \ +/- \ \ \sqrt{b^2 - 4ac} }{2a} \\\\t = \frac{-(-20) \ \ +/- \ \ \sqrt{(-20)^2 - 4(1.8\times 25)} }{2(1.8)}\\\\t = 9.67s, \ \ or \ 1.43 \ s[/tex]

The value of t that will give 25 m upwards using the motion model is 1.43 s;

h = 20(1.43) - 1.86(1.43)² = 25 m

The velocity of the rock at 25 m upwards (t = 1.43 s) is calculated as follows;

[tex]v = \frac{dh}{dt} = 20 -3.72t\\\\v = 20 - 3.72(1.43) \\\\v = 14.68 \ m/s[/tex]

The velocity of the rock at 25 m downwards (t = 9.67 s) is calculated as;

[tex]v = \frac{dh}{dt} = 20 - 3.72t\\\\v = 20 - 3.72(9.67)\\\\v = -15.97 \ m/s[/tex]

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