Question

"how long would it take for a ball dropped from the top of a 400-foot building to hit the ground? round your answer to two decimal places."

Answer 1

Hello

Let's convert first the height from foot to meters:
$S=400~ft=121.92~m$
This is the space covered by the ball in its accelerated motion, where the acceleration is the gravitational acceleration:
$a=g=9.81~m/s^2$
The law of motion is given by
$S= \frac{1}{2}at^2$
Re-arranging, we can get the total time of the motion:
$t= \sqrt{ \frac{2S}{a} }=4.99~s$
which is the time the ball takes to hit the ground.

Answer 2

Given:
s = 400 ft, the height traveled

Use
g = 32.2 ft/s², acceleration due to gravity
Ignore air resistance.

Use the formula
s = ut +(1/2)gt²
where
u = initial velocity,
t  = time, sec

Because the ball was dropped, the initial velocity is zero.
Therefore
400 = (1/2)*32.2*t² =16.1*t²
t = √(400/16.1) = 4.98 s (approximately 5 s)

Answer: 5 s