# A projectile is shot from the ground at a velocity of #1 m/s# and at an angle of #(pi)/2#. How long will it take for the projectile to land?

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To find the time it takes for the projectile to land, we can use the equation for the horizontal motion of a projectile: ( t = \frac{2v_0\sin(\theta)}{g} ), where ( v_0 ) is the initial velocity, ( \theta ) is the angle of projection, and ( g ) is the acceleration due to gravity (approximately 9.8 m/s²). Plugging in the given values, we get: ( t = \frac{2 \times 1 \times \sin(\pi/2)}{9.8} = \frac{2}{9.8} ) seconds. Solving this equation gives us ( t = 0.204 ) seconds. Therefore, it will take approximately 0.204 seconds for the projectile to land.

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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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