If a projectile is shot at an angle of #pi/6# and at a velocity of #18 m/s#, when will it reach its maximum height??
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To find the time it takes for a projectile to reach its maximum height, you can use the formula: ( t = \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 )). Plug in the values: ( v_0 = 18 , \text{m/s} ) and ( \theta = \pi/6 ). Calculate ( t ).
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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.
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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