# Question #79d47

##### 1 Answer

#### Answer:

#### Explanation:

The first thing to do here is to figure out the time needed for the rock to *hit the ground*.

Your tool of choice here is the equation

#color(blue)(ul(color(black)(h = v_0 * t + 1/2 * g * t^2)))#

Here

#h# is the distance covered by the object in a given time#t# #v_0# is the initial velocity of the rock#t# is the time of flight#g# is the gravitational acceleration, usually given as#"9.81 m s"^(-2)#

In your case, the object starts from rest, so

#v_0 = "0 m s"^(-1)#

This means that you will have

#h = 1/2 * g * t^2#

Rearrange to solve for

#t = sqrt((2 * h)/g)#

Plug in your values to find

#t = sqrt( (2 * 28.5 color(red)(cancel(color(black)("m"))))/(9.81color(red)(cancel(color(black)("m")))"s"^(-2))) = "2.41 s"#

So, you know that it takes **before** reaching the

#t_"needed" = "2.41 s" - "1.2 s" = "1.21 s"#

Use this value in the above equation and solve for **in the first**

#h_"1.21 s" = 1/2 * 9.81 color(red)(cancel(color(black)("m"))) color(red)(cancel(color(black)("s"^(-2)))) * (1.21)^2 color(red)(cancel(color(black)("s"^2)))#

#h_"1.21 s" = "7.18 m"#

Therefore, you can say that **after** it began to fall and **before** hitting the ground, the rock was

#color(darkgreen)(ul(color(black)(h_ "above ground" = "28.5 m" - "7.18 m" = "21.3 m"))) -># above the ground