Ironman steps from the top of a tall building. He falls freely from rest to the ground a distance of h. He falls a distance of h/ 3 in the last interval of time of 1.1 s of his fall.

Question

Hint: First, compute the velocity when Ironman reaches the height equal to the distance fallen. This requires that you do the following: define origin as the bottom of the building. Then use x-x0 = -v0*(t-t0)-(1/2)g(t-t0)^2 where x=0 and x0= (distance fallen) and t-t0 is the time interval given. In this formulation, you are going to get magnitude of v0 since you already inserted the sign.

You then insert v0 that you just calculated into the kinematic equation that involves v, g, and displacement (v^2-v0^2 = 2g(height-(distance fallen)), but now v (which is the final velocity is v0 from above) and v0 in this case is the velocity that the Ironman has when he begins to fall, which is 0.

This gives a quadratic equation for height h, and you will need to use the binomial equation to solve for h. Choose the larger of the two solutions.

Answer 1

Theres a simpler way to do this, since there is zero initial velocity. We might as well use the frame of reference , the top of the building is our origin. And let down movement be positive.

Since there is zero initial velocity. We might as well use the top of the building as our origin. And let down movement be positive , so he accelerating in the same direction as his motion.

The equation for the distance the ironman moved
is y = 1/2 gt^2

Let h = total height, and T be total time.
h = 1/2 gT^2

Now it tells us that when T - 1.1 sec the ironman is h/3 units of distance from the ground. This is the same thing as saying he covered h*2/3 in T-1.1 seconds.
2/3 *h = 1/2 g ( T - 1.1)^2

now its simpler if we substitute 1/2gT^2 for h. The other way is more tedious because of the square rooting.

2/3 * (1/2 gT^2) = 1/2 g ( T - 1.1)^2

2/3 T^2 = (T - 1.1)^2

2/3 T^2= T^2 - 2T*1.1 + 1.1^2

0 = 1/3 T^2 - 2.2T + 1.21
0 = T^2 -6.6 T + 3.63

This gives us two solutions T= 5.9944 ,0.6056 approximately
The second solution we reject because we want T - 1.1 , and that will make it negative.

So that's T, and using 9.8 meters /s^2 for g we have
h = 1/2 9.8 (5.9944)^2 meters
and height is roughly 176 meters high to the nearest meter