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

1. 9.04 s

We can find the time taken for the steel ball to reach the ground by using the SUVAT equation:

`d = ut + (1)/(2)gt^2`

where

d = 400 m is the distance

u = 0 is the initial velocity of the ball

g = 9.8 m/s^2 is the acceleration of gravity

t is the time

Solving the formula for t, we find the time taken for the ball to reach the ground:

`t=\sqrt{(2d)/(g)}=\sqrt{(2(400))/(9.8)}=9.04 s`

2. 88.6 m/s

The final velocity of the ball before it reaches the ground can be found by using the equation

v = u + gt

where

u is the initial velocity

g is the acceleration of gravity

t is the time

Here we have

u = 0

g = 9.8 m/s^2

Substituting the time of flight, t = 9.04 s, we find the final velocity:

`v=0+(9.8)(9.04)=88.6 m/s`