# If a projectile is shot at an angle of #pi/12# and at a velocity of #1 m/s#, when will it reach its maximum height??

The maximum height is

The highest point is determined by

Thus,

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The projectile will reach its maximum height at the time when its vertical velocity component becomes zero. Using the equations of motion, the time to reach maximum height ( t_{\text{max}} ) is given by ( 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. Substituting ( V_0 = 1 ) m/s, ( \theta = \frac{\pi}{12} ), and ( g = 9.8 ) m/s(^2), we can calculate ( t_{\text{max}} ).

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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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