# If a projectile is shot at an angle of #pi/6# and at a velocity of #5 m/s#, when will it reach its maximum height??

The velocity's initial upward component is

The projectile's vertical component of velocity is zero at its highest point. Why is this the case? If the projectile is still moving upward, it can still travel a greater distance. If it is moving downward, it must have descended from a location higher than where it is now.

We've got

That concludes it.

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The projectile will reach its maximum height halfway through its flight time. The time to reach maximum height can be calculated using the formula: ( t_{\text{max}} = \frac{V_y}{g} ), where ( V_y ) is the vertical component of the initial velocity and ( g ) is the acceleration due to gravity (approximately ( 9.8 , \text{m/s}^2 )). Substituting the given values, the time to reach maximum height is approximately ( t_{\text{max}} = \frac{5 \sin(\pi/6)}{9.8} ). Calculating this gives the time it takes to reach maximum height.

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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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