Question #a307c
1 Answer
The total time of its fall is equal to
The total distance covered is equal to
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
Use the value of
#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.
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,
#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
#{ (h/2 = g * t_1 + 5 = 10t_1 + 5), (h/2 = 5 * t_1^2) :}#
Now solve for
#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
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")#