Question

What is the time required for an object starting at rest to fall freely 500 meters near earth surface

Answer 1

Without air resistance . . .

           Distance  =  1/2 g T²

               500 m  =  (1/2) (9.8 m/s²) (T²)

               500 m  =  (4.9 m/s²) (T²)

Divide each side by   (4.9 m/s²)  :

                  T²  =  (500 m) / (4.9 m/s²)

                       =     102.04 s²

                   T  =  √ (102.04 s²)

                       =       10.1 seconds       

Answer 2

The time required for this object to fall freely is 10.10 seconds.

Given the following data:

• Initial velocity, u = 0 m/s (since the object is starting from rest).

• Distance, S = 500 meters.

We know that acceleration due to gravity (a) for an object in free fall is equal to 9.8 meter per seconds square.

To find the time required for this object to fall freely, we  would use the second equation of motion.

Mathematically, the second equation of motion is given by the formula;

$S = ut + \frac{1}{2} at^2$

Where:

• S is the displacement or distance covered.

• u is the initial velocity.

• a is the acceleration.

• t is the time measured in seconds.

Substituting the values into the formula, we have;

$500 = 0(t) + \frac{1}{2} (9.8)t^2\\\\500 = 4.9t^2\\\\t^2 = \frac{500}{4.9}\\\\t^2 = 102.04\\\\t = \sqrt{102.04}$

Time, t = 10.10 seconds

Therefore, the time required for this object to fall freely is 10.10 seconds.

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