# A projectile is shot from the ground at an angle of #pi/6 # and a speed of #15 m/s#. Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?

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To find the projectile's distance from the starting point at maximum height, use the formula: [ \text{Horizontal distance} = \frac{v_0^2 \sin(2\theta)}{g} ] where ( v_0 ) is the initial speed, ( \theta ) is the launch angle, and ( g ) is the acceleration due to gravity. Plugging in the values: ( v_0 = 15 , \text{m/s} ), ( \theta = \pi/6 ), and ( g = 9.8 , \text{m/s}^2 ), calculate the horizontal distance.

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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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