# A projectile is shot at an angle of #pi/3 # and a velocity of # 24 m/s#. How far away will the projectile land?

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To find the horizontal distance the projectile will travel, you can use the horizontal component of the initial velocity and the time of flight. First, calculate the horizontal component of the initial velocity using trigonometry:

[ v_{0x} = v_0 \cdot \cos(\theta) ]

[ v_{0x} = 24 \cdot \cos\left(\frac{\pi}{3}\right) ]

[ v_{0x} = 24 \cdot \frac{1}{2} ]

[ v_{0x} = 12 , \text{m/s} ]

Next, find the time of flight using the vertical motion equation:

[ y = v_{0y}t - \frac{1}{2}gt^2 ]

Since the projectile lands at the same height it was launched, the final vertical position (y) is 0. Also, the initial vertical velocity (v_{0y}) can be found using the sine of the launch angle:

[ v_{0y} = v_0 \cdot \sin(\theta) ]

[ v_{0y} = 24 \cdot \sin\left(\frac{\pi}{3}\right) ]

[ v_{0y} = 24 \cdot \frac{\sqrt{3}}{2} ]

[ v_{0y} = 12\sqrt{3} , \text{m/s} ]

Substitute the known values into the equation:

[ 0 = (12\sqrt{3})t - \frac{1}{2}(9.8)t^2 ]

[ 0 = 12\sqrt{3}t - 4.9t^2 ]

[ 4.9t^2 = 12\sqrt{3}t ]

[ t = \frac{12\sqrt{3}}{4.9} ]

[ t \approx 2.76 , \text{s} ]

Finally, find the horizontal distance traveled using the time of flight and horizontal velocity:

[ d = v_{0x} \cdot t ]

[ d = 12 \cdot 2.76 ]

[ d \approx 33.12 , \text{m} ]

Therefore, the projectile will land approximately 33.12 meters away.

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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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