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/ 4 in the last interval of time of 1.1s 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-vO^2 = 2g(height-(distance fallen)), but now v (which is the final velocity is vo from above) and vO 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

College Physics
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Chapter1: Units, Trigonometry. And Vectors
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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 \( \frac{h}{4} \) in the last interval of time of 1.1 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 the 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 the 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 \( (v^2-v0^2 = 2g(\text{height-distance fallen})) \), 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.

---

**Part A**

What is the height \( h \) of the building?

*Express your answer using two significant figures.*

\[ h = \, \_\_\_\_ \, \text{m} \]

---

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The interactive box allows users to input their height calculation and submit it for validation.
Transcribed Image Text: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 \( \frac{h}{4} \) in the last interval of time of 1.1 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 the 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 the 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 \( (v^2-v0^2 = 2g(\text{height-distance fallen})) \), 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. --- **Part A** What is the height \( h \) of the building? *Express your answer using two significant figures.* \[ h = \, \_\_\_\_ \, \text{m} \] --- **Buttons and Options**: - Submit - Request Answer - Provide Feedback The interactive box allows users to input their height calculation and submit it for validation.
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