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An object is thrown vertically from a height of #2 m# at #21 m/s#. How long will it take for the object to hit the ground?

Answer 1

William Audy

#0.092 s#

Let's calculate the speed at which it will land by using

#v^2=u^2 +2gs#

Given, #u=21ms^-1,s=2m#

So, #v^2=21×21 +2×9.8×2#

So, #v=21.92ms^-1#

Now,if it took time #t# in gaining that much velocity we can use, #v=u+g t#

So, #21.92=21 + 9.8t#

Or, #t=0.092 s#

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Answer 2

Layla Ahmed

To find the time it takes for the object to hit the ground, we can use the kinematic equation:

[ \text{time} = \sqrt{\frac{2 \times \text{height}}{g}} ]

Where:

( \text{height} = 2 ) meters (initial height)
( g ) is the acceleration due to gravity, approximately ( 9.81 , \text{m/s}^2 )

Plugging in the values:

[ \text{time} = \sqrt{\frac{2 \times 2}{9.81}} ]

[ \text{time} \approx \sqrt{\frac{4}{9.81}} ]

[ \text{time} \approx \sqrt{0.407} ]

[ \text{time} \approx 0.638 , \text{s} ]

So, it will take approximately 0.638 seconds for the object to hit the ground.

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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.