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  

they should be similar or the same

mass and volume are the answers you are looking for