An object is thrown vertically from a height of #12 m# at # 2 m/s#. How long will it take for the object to hit the ground?

Answer 1

I found #1.8s#

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Answer 2

Using the equation of motion for vertically thrown objects, ( h = ut + \frac{1}{2}gt^2 ), where ( h ) is the initial height (12 m), ( u ) is the initial velocity (2 m/s), ( g ) is the acceleration due to gravity (9.8 m/s²), and ( t ) is the time taken. Rearranging the equation to solve for ( t ), we get:

( h = ut + \frac{1}{2}gt^2 )

( 12 = 2t + \frac{1}{2}(9.8)t^2 )

( 9.8t^2 + 2t - 12 = 0 )

Using the quadratic formula, ( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ), where ( a = 9.8 ), ( b = 2 ), and ( c = -12 ), we find:

( t = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 9.8 \cdot (-12)}}{2 \cdot 9.8} )

( t = \frac{-2 \pm \sqrt{4 + 470.4}}{19.6} )

( t = \frac{-2 \pm \sqrt{474.4}}{19.6} )

( t = \frac{-2 \pm 21.79}{19.6} )

( t = \frac{19.79}{19.6} ) or ( t = \frac{-23.79}{19.6} )

Since time cannot be negative, we take the positive value:

( t \approx 1.01 ) seconds.

So, it will take approximately 1.01 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.

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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