# A projectile is shot at an angle of #(5pi)/12 # and a velocity of # 2 m/s#. How far away will the projectile land?

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To find the horizontal distance traveled by the projectile, you can use the formula:

[ \text{Range} = \frac{v^2 \times \sin(2\theta)}{g} ]

Where:

- ( v ) = initial velocity of the projectile (m/s)
- ( \theta ) = launch angle (radians)
- ( g ) = acceleration due to gravity (approximately ( 9.8 , \text{m/s}^2 ))

Given:

- ( v = 2 , \text{m/s} )
- ( \theta = \frac{5\pi}{12} )

Plug the values into the formula:

[ \text{Range} = \frac{2^2 \times \sin(2 \times \frac{5\pi}{12})}{9.8} ]

[ \text{Range} \approx 0.863 , \text{m} ]

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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.

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