Ironman steps from the top of a tall building. He falls freely from rest to the

Question

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 1.2 s of his fall. 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.

Explanation / Answer

Let the height of the building = h

From the hint, we have

x - x0 = - v0 (t - t0) - (1/2) g (t - t0)^2

- 3 =  - v0 (-1.2) - (1/2) g (-1.2)^2

- 3 = 1.2 v0 - (1/2) g (-1.2)^2

by solving we get, v0 = 5.88 m/s

and again we have, v^2 - v0^2 = 2 g (h - 3)

Here, v = 5.88 m/s   and v0 means the initial velocity of the man when he falls i.e. zero.

          (5.88)^2 - 0^2 = 2 * 9.8 (h - 3)                   

          (5.88)^2 = 2 * 9.8 (h - 3)               

by solving the above we get, h = 4.764 m

So the height of the building, h = 4.764 m    

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