# 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

*total distance* of the fall;

**enters the bottom half** of the total distance;

**1 second**;

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 *negative*, the only feasible solution is

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")#