Question

An object in free fall near the surface of the earth accelerates at a rate of 78979.4 mi/hr What is the rate of acceleration for a free falling object near the surface of the earth in ft/s and m/s Press you used to arrive at this answer. r. Approximate the number of exhalations an average person takes in one lifetime?

Answer 1

Answer:

32.176 ft/s²

9.807 m/s²

Explanation:

Data provided in the question:

Acceleration of the free falling object = 78979.4 mi/hr²

a) In ft/s

Now,

1 mi = 5280 foot

and,

1 hour = 3600 seconds

thus,

78979.4 mi/hr = $\frac{\textup{78979.4}\times\textup{5280}}{\textup{1}\times\textup{3600}^2}$

or

78979.4 mi/hr = 32.176 ft/s²             ...............(1)

b) In m/s²

Now,

1 foot = 0.3048 m

thus,

we have from 1

32.17 ft/s² =  $\frac{\textup{32.17}\times\textup{0.3048}}{\textup{1}\times\textup{1}^2}$

or

78979.4 mi/hr = 32.176 ft/s²  = 9.807 m/s²