Arock is projected upward from the surface of the moon, at time t = 0.0 s, with a velocity of the acceleration due to gravity at the surface of the moon is when does the rock reach a height of 180 m above the moon's surface?
17 s
12 s
30 s
23 s
7.5 s

7.5 seconds

Given u= initial velocity=30m/sec

          a= acceleration on moon=1.62m/s2

           h=180m

Using the equation

h = u(t) - 1/2gt² {where g = 1.62 m/s²}   and u = 30 m/s

180 = 30t - (0.5)(1.62)t²  

0 = 0.81t² - 30t +180  

solving the quadratic equation for the 2 t-values:  

t = 7.5 s or t = 30 s

Therefore; the time taken to reach a height of 180 m upwards is 7.5 seconds

it is the second option.

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y