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

Explanation:

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

$y=y_{o}+V_{o}t+\frac{1}{2}gt^{2}$ (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:

$0=y_{o}+\frac{1}{2}gt^{2}$ (3)  

Isolating $t$:

$t=\sqrt{\frac{-2y_{o}}{g}}$ (4)  

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