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?
0.56518s
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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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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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