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

Answer 1

0.56518s

For a distance, #d#, an initial velocity, #v_0#, an acceleration of #a#, and a time of #t# #color(white)("XXX")d=v_0t+at^2#
In this case #color(white)("XXX")d=-2 " m"# (2 meters below the height from with it was thrown)If it was dropped and not thrown up there would be no negative sign #color(white)("XXX")v_0= 2 " m"/"sec"# #color(white)("XXX")a=-9.8 " m"/(sec"^2)# (standard gravity assumed)
So we have #color(white)("XXX")-2.0" m" = 2.0t " m" + (-9.8t^2)" m"#
#color(white)("XXX")98t^2-20t-20=0#
Using the quadratic formula: #color(white)("XXX")t=(20+-sqrt(20^2-4(98)(-20)))/(2(98))#
#color(white)("XXX")t = 0.56518s#
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Answer 2

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

[ s = ut + \frac{1}{2}at^2 ]

Where:

  • ( s ) = displacement (change in height)
  • ( u ) = initial velocity
  • ( a ) = acceleration due to gravity (-9.8 m/s^2)
  • ( t ) = time

Rearranging the equation and plugging in the known values:

[ -2 = (2)t + \frac{1}{2}(-9.8)t^2 ]

This simplifies to a quadratic equation:

[ 4.9t^2 - 2t - 2 = 0 ]

Using the quadratic formula, we find:

[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ]

[ t = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(4.9)(-2)}}{2(4.9)} ]

[ t = \frac{2 \pm \sqrt{4 + 39.2}}{9.8} ]

[ t \approx \frac{2 \pm \sqrt{43.2}}{9.8} ]

Solving for ( t ):

[ t \approx \frac{2 \pm 6.57}{9.8} ]

[ t \approx \frac{8.57}{9.8} ] (ignoring the negative root as time can't be negative)

[ t \approx 0.875 ]

So, it will take approximately 0.875 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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