If an object is dropped, how fast will it be moving after #16 s#?

Answer 1
Theoretical: #v=u+at#, where:
We will take #a=9.81ms^-2#
#v=0+16(9.81)=156.96ms^-1~~157ms^-1#

Realistic: The speed will depend on the shape of the object and surface area (large drag force or small drag force), height it is dropped from (to allow for a 16s fall), environment (different mediums will have different drag forces for the same object), how high the object is (higher up you go, the smaller the drag force but the smaller the acceleration due to gravity).

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Answer 2

The speed of an object dropped near the surface of the Earth can be calculated using the formula for free fall: ( v = gt ), where ( v ) is the final velocity, ( g ) is the acceleration due to gravity (approximately ( 9.8 , \text{m/s}^2 )), and ( t ) is the time in seconds.

Plugging in the values, ( v = (9.8 , \text{m/s}^2)(16 , \text{s}) ), we get:

( v = 156.8 , \text{m/s} ).

So, the object will be moving at approximately ( 156.8 , \text{m/s} ) after ( 16 , \text{s} ) of free fall.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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