# If a projectile is shot at an angle of #pi/6# and at a velocity of #24 m/s#, when will it reach its maximum height??

Note that

will be 0 at maximum height

But at maximum height

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The projectile will reach its maximum height at the time given by the formula: (t_{\text{max}} = \frac{V_0 \sin(\theta)}{g}), where (V_0) is the initial velocity, (\theta) is the launch angle, and (g) is the acceleration due to gravity (approximately (9.8 , \text{m/s}^2)). Substituting the given values: (t_{\text{max}} = \frac{24 \sin(\pi/6)}{9.8}). Calculate to find the time.

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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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