Question

The Sears Tower is nearly 400 m high. How long would it take a steel ball to reach the ground if dropped on the top? What will be it's velocity the moment for it touches the ground?

Answer 1

Answers:

a) 9.035 s

b) -88.543 m/s

Step-by-step explanation:

The described situation is related to vertical motion (especifically free fall) and the equations that will be useful are:

`y=y_(o)+V_(o)t+(1)/(2)gt^(2)` (1)

`V=V_(o)+gt` (2)

Where:

`y=0` is the final height of the steel ball

`y_(o)=400 m` is the initial height of the steel ball

`V_(o)=0` is the initial velocity of the steel ball (it was dropped)

`V` is the final velocity of the steel ball

`t` is the time it takes to the steel ball to reach the ground

`g=-9.8 m/s^(2)` is the acceleration due to gravity

Knowing this, let's begin with the answers:

a) Time it takes the steel ball to reach the ground

We will use equation (1) with the conditions listed above:

`0=y_(o)+(1)/(2)gt^(2)` (3)

Isolating
`t`:

`t=\sqrt{(-2y_(o))/(g)}` (4)

`t=\sqrt{(-2(400 m))/(-9.8 m/s^(2))}` (5)

`t=9.035 s` (6)

b) Final velocity of the steel ball

We will use equation (2) with the conditions explained above and the calculaated time:

`V=gt` (7)

`V=(-9.8 m/s^(2))(9.035 s)` (8)

`V=-88.543 m/s` (9) The negative sign indicates the direction of the velocity is downwards