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3 years ago

9

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 b

e it’s velocity the moment for it touches the ground?

1 answer:

Damm [24]3 years ago

4 0

1. 9.04 s

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

$d = ut + \frac{1}{2}gt^2$

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:

$t=\sqrt{\frac{2d}{g}}=\sqrt{\frac{2(400)}{9.8}}=9.04 s$

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:

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

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