An object falls a distance h from rest. If it travels .50h in the last 1.00s, find (a) the time and (b) the height of its fall?

1 Answer
Sep 7, 2015

Answer:

Height: #"57.1 m"#
Time: #"3.41 s"#

Explanation:

So, you know that your objects starts falling from rest from a height #h#. Moreover, you know that it covered a distance of #0.5h# in the last #"1.00 s"# of its motion.

Since this distance of #0.5h# was covered 1.00 seconds before reaching the ground, you can say that the distance it covered in the first part of its motion, from top position to #0.5h# is

#h_"first" = h - 0.5h = 0.5h#

So, it covered bottom half of its motion in one second, which means that you can write

#0.5h = v * "1 s" + 1/2 * g * "1 s"^2" "#, where

#v# - the velocity of the object after it travelled the first half of its motion.

#0.5h = v * 1 + 1/2 * 9.8 * 1 = v + 4.9#

Now focus on the first half of its motion. You can find a second relationship between #v# and #h# by writing

#v^2 = underbrace(v_0^2)_(color(blue)(=0)) + 2 * g * h_"first"#

#v^2 = 2 * g * 0.5h = g * h = 9.8 * h#

So, you have

#{(0.5h = v + 4.9), (v^2 = 9.8 * h):}#

Use the first equation to find #h# as a function of #v#, then plug that into the second equation to find #h#

#h = (v + 4.9)/0.5 = 2v + 9.8#

#v^2 - 9.8 * (2v + 9.8) = 0#

#v^2 - 19.6v - 96.04 = 0#

Use the quadratic formula to find the two solutions to this quadratic equation

#v_(1,2) = (-(-19.6) +- sqrt((-19.6)^2 - 4 * 1 * 96.04))/(2 * 1)#

#v_(1,2) = (19.6 +- sqrt(768.32))/2#

The negative solution has no physical significance in this context, which means that #v# will be

#v = (19.6 + 27.72)/2 = "23.66 m/s"#

Use this value to find #h#

#h = 2 * 23.66 + 9.8 = color(green)("57.1 m")#

To find the total time of flight, use

#h = underbrace(v_0)_(color(blue)(=0)) + 1/2 * g * t_"total"^2#

#t_"total" = sqrt((2 * h)/g) = sqrt((2 * 57.12color(red)(cancel(color(black)("m"))))/(9.8color(red)(cancel(color(black)("m")))/"s"^2)) = color(green)("3.41 s")#