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

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  

energy can be converted from one form to another, but it cannot be created or destroyed in ordinary chemical or physical means. do not understand what you are asking but hope this