Question #79d47

1 Answer
Mar 4, 2017

Answer:

#"21.3 m"#

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#

#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 #"2.41 s"# for the rock to hit the ground. Use this value to determine the time spent falling before reaching the #"1.2-s"# mark

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

Use this value in the above equation and solve for #h#, the distance covered by the rock in the first #"1.21 s"# of falling

#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 #"1.21 s"# after it began to fall and #"1.2 s"# 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