Question

The John Hancock Center in Chicago is the tallest building in the United States in which there are residential apartments. The Hancock Center is 343 m tall. Suppose a resident accidentally causes a chunk of ice to fall from the roof. What would be the velocity of the ice as it hits the ground

Answer 1

Answer:

Explanation:

The chunk of ice will fall with acceleration of 9.8 m /s²

initial velocity u = 0 ,

v² = u² + 2 gH

H = 343 m

g = 9.8 m /s²

v is final velocity

v² = 0 + 2 x 9.8 x 343

v² = 6723

v = 82 m /s approx .

Answer 2

The velocity of the the chunk of ice as it hits the ground is 82m/s.

Given the data in the question;

• Height of John Hancock Center; $h = 343m$

Before the chunk of ice to fell off the roof, its was initially at rest, so

• Initial velocity; $u = 0$

Final velocity as the ice hits the ground; $v = \ ?$

To determine the final velocity, we use the third equation of motion:

$v^2 = u^2 + 2as$

Where v is final velocity, u is the initial velocity, s is distance or height of the building and a is acceleration due to gravity{ since the ice will be under gravity as it falls, ($a = g = 9.8m/s^2$) }

We substitute our values into the equation

$v^2 = [0^ 2] + [ 2\ *\ 9.8m/s^2\ *\ 343m]\\\\v^2 = 2\ *\ 9.8m/s^2\ *\ 343m\\\\v^2 = 6722.8m^2/s^2\\\\v = \sqrt{6722.8m^2/s^2}\\\\v = 81.99m/s\\\\v = 82m/s$

Therefore, the velocity of the the chunk of ice as it hits the ground is 82m/s.

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