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"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."

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skyluke89
Hello

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

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