Question #a307c

1 Answer
Jun 11, 2015

Answer:

The total time of its fall is equal to #t_"total" = (2+sqrt(2))"s"#.
The total distance covered is equal to #h = 10(3 + 2sqrt(2))"m"#.

Explanation:

I think that you can go about solving this one in others ways as well, but here's the approach I used.

You know that the object covered half of its total path in 1 second. Moreover, it was the last second of its fall. This means that the distance it covered is actually the bottom half. You can write

#underbrace(h/2)_(color(blue)("second half of the fall")) = v_1t + 1/2g * t^2#, where

#h# - the total distance of the fall;
#v_1# - the speed with which it enters the bottom half of the total distance;
#t# - the time it took for this distance to be covered - in your case, #t# is equal to 1 second;
#g# - the gravitational acceleration, which I'll use as #"10 m s"^(-2)#

Use the value of #t# to determine a relationship between #h/2# and #v_1#.

#h/2 = v_1 * 1 + 1/2 * 10 * 1^2 = v_1 + 5# #" "color(blue)((1))#

Now focus on the first half of the fall. You know that your object covers the same distance, i.e. #h/2#, in a time I'll label as #t_1#.

Once again, you can write

#underbrace(h/2)_(color(blue)("first half of the fall")) = v_0 * t_1 + 1/2g * t_1^2#

Since the object starts from a stationary position, #v_0 = 0#. This will get you

#h/2 = 1/2 * 10 * t_1^2 = 5 * t_1^2# #" "color(blue)((2))#

For the first half of the fall, you can also write this equation

#v_1 = v_0 + g * t_1 <=> v_1 = g * t_1#

Use this identity to replace #v_1# in equation #color(blue)((1))# to get

#{ (h/2 = g * t_1 + 5 = 10t_1 + 5), (h/2 = 5 * t_1^2) :}#

Now solve for #t_1#.

#5t_1^2 = 10t_1 + 5 <=> t_1^2 - 2t_1 -1 = 0#

This quadratic has two solutions

#t_(1)^' = (2 + sqrt(4 + 4))/2 = 1 + sqrt(2)#

and

#t_(1)^('') = (2-sqrt(4 + 4))/2 = 1-sqrt(2)#

Since #t_1^('')# is negative, the only feasible solution is #t_1^(') = 1 + sqrt(2)#.

This means that the total time of the fall was

#t_"total" = t_1 + t = 1 + sqrt(2) + 1 = color(green)(2 + sqrt(2)" seconds")#

The total distance covered was

#h = 1/2 * g * t_"total"^2 = 5 * (2 + sqrt(2))^2#

#h = 5 * (4 + 4sqrt(2) + 2) = 30 + 20sqrt(2) = color(green)(10(3 + 2sqrt(2)" meters")#