# A projectile is shot from the ground at a velocity of #19 m/s# at an angle of #pi/4#. How long will it take for the projectile to land?

The time is

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To find the time it takes for the projectile to land, you can use the equation:

[ t = \frac{2v_0\sin(\theta)}{g} ]

where:

- ( t ) is the time of flight,
- ( v_0 ) is the initial velocity (19 m/s),
- ( \theta ) is the launch angle (π/4 or 45 degrees), and
- ( g ) is the acceleration due to gravity (approximately 9.8 m/s²).

Substituting the given values:

[ t = \frac{2 \times 19 \times \sin(\frac{\pi}{4})}{9.8} ]

[ t = \frac{38 \times \frac{\sqrt{2}}{2}}{9.8} ]

[ t = \frac{19\sqrt{2}}{9.8} ]

[ t \approx 2.02 , \text{seconds} ]

So, it will take approximately 2.02 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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